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UK 1983

#UK1983#QED SYSTEMS#ACTION#STRATEGY#DRAGON32#ATLANTIS#DRAGON DOMAIN#CARDS#AWARI#DRAXIT#LINK FOUR#PHOTONS#PHANTOMS#MICROTHELLO

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[Toy Bonnie Among Us headcanons, requested by a friend.]

Toy Bonnie Among Us headcanons! First picture is with full body pictures, while the second is without. I thought it’d be fun to include System Error, too which really brought this moodboard together.

My headcanons, since it’s a little hard to read on the moodboard, are as follows:

Really likes being imposter

Always manages to get an alibi

Good liar like bro teach me

VENT GOBLIN

Can help a ton or cause chaos

Hangs out in the hallways

Basically 3rd imposter

Fake surprise

The song included is “The Other Side of Paradise” by Glass Animals.

#tw eyestrain#qed#interstellar's moodboard#fnaf moodboard#fnaf 2#fnaf 2 toy bonnie#toy bonnie#five nights at freddy's#among us moodboard#among us#headcanons#among us headcanons#system error toy bonnie#red#blue#moodboards#moodboard#DANG IT I MISSPELLED SURPRISE#f#personal moodboard#glass animals#song lyrics

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Circumscribed Rectangle

Calculates the side length and area of the regular polygon inscribed to a circle.

[1-10] /72Disp-Num

[1] 2021/04/18 10:50 Male / 60 years old level or over / A retired person / Very /

Purpose of use

Needed a simple calculator rather than proofs of mathematical formula.

Comment/Request

Could use a visual representation of actual values applied to shapes

[2] 2021/02/18 01:03 Male / 20 years old level / An engineer / Useful /

Purpose of use

3D modeling for use in finite element analysis of a ring shape.

[3] 2021/02/16 12:01 Male / Under 20 years old / Elementary school/ Junior high-school student / Very /

Purpose of use

calculate the approximate area of my massive cock ring

[4] 2021/01/29 08:11 Male / Under 20 years old / High-school/ University/ Grad student / Useful /

Purpose of use

Dodecagon calculations

[5] 2021/01/15 01:19 Male / 60 years old level or over / A retired person / Very /

Purpose of use

To construct a heptagon poker table (Hey- anyone can do a octagon!)

[6] 2020/12/22 06:51 Male / 30 years old level / An engineer / Useful /

Purpose of use

Creating an algorithm to calculate lead-in on a CNC mill interpolating a circle.

[7] 2020/09/05 03:49 Male / 20 years old level / An engineer / Very /

Purpose of use

create a grid of points with minimum/maximum distance specified

[8] 2020/06/09 16:08 Male / 20 years old level / High-school/ University/ Grad student / Useful /

Purpose of use

Estimating the value of pi

[9] 2020/05/07 18:40 Male / 60 years old level or over / An engineer / Very /

Purpose of use

Various calculations

[10] 2020/03/25 11:54 Male / 60 years old level or over / An engineer / Very /

Purpose of use

planting a circular walnut grove comprised of concentric circular plantings on approximate 40' spacing

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Circumscribed Angle Definition

Circumscribed Rectangle, Bounding Box. (Coordinate Geometry) - Math ...

Now using the Lagrange Multipliers technique. The circumscribed rectangle has the side dimensions #(a+b)# and #(c+d)# so the sough area is #(a+b)(c+d)# The restrictions are. Inscribed rectangle and less than the area of a circumscribed rectangle. The Mean Value Theorem for Integrals states that somewhere “between” the inscribed and circumscribed rectangles there is a rectangle whose area is precisely equal to the area of the region under the curve.

Recall from the Law of Sines that any triangle (triangle,ABC) has a common ratio of sides to sines of opposite angles, namely

Circumscribed Angle Definition

[nonumber
frac{a}{sin;A} ~=~ frac{b}{sin;B} ~=~ frac{c}{sin;C} ~.
]

This common ratio has a geometric meaning: it is the diameter (i.e. twice the radius) of the unique circle in which (triangle,ABC) can be inscribed, called the circumscribed circle of the triangle. Before proving this, we need to review some elementary geometry.

An inscribed angle of a circle is an angle whose vertex is a point (A) on the circle and whose sides are line segments (called chords) from (A) to two other points on the circle. In Figure 2.5.1(b), (angle,A) is an inscribed angle that intercepts the arc (overparen{BC} ). We state here without proof a useful relation between inscribed and central angles:

Theorem 2.4

If an inscribed angle (angle,A) and a central angle (angle,O) intercept the same arc, then (angle,A = frac{1}{2},angle,O, ). Thus, inscribed angles which intercept the same arc are equal.

Figure 2.5.1(c) shows two inscribed angles, (angle,A) and (angle,D ), which intercept the same arc (overparen{BC}) as the central angle (angle,O ), and hence (angle,A = angle,D = frac{1}{2},angle,O) (so (;angle,O = 2,angle,A = 2,angle,D,) ).

We will now prove our assertion about the common ratio in the Law of Sines:

Theorem 2.5

For any triangle (triangle,ABC ), the radius (R) of its circumscribed circle is given by:

[2,R ~=~ frac{a}{sin;A} ~=~ frac{b}{sin;B} ~=~ frac{c}{sin;C}label{2.35}]

Note: For a circle of diameter (1 ), this means (a=sin;A ), (b=sin;B ), and (c=sin;C ).)
To prove this, let (O) be the center of the circumscribed circle for a triangle (triangle,ABC ). Then (O) can be either inside, outside, or on the triangle, as in Figure 2.5.2 below. In the first two cases, draw a perpendicular line segment from (O) to (overline{AB}) at the point (D ).

The radii (overline{OA}) and (overline{OB}) have the same length (R ), so (triangle,AOB) is an isosceles triangle. Thus, from elementary geometry we know that (overline{OD}) bisects both the angle (angle,AOB) and the side (overline{AB} ). So (angle,AOD = frac{1}{2},angle,AOB) and (AD = frac{c}{2} ). But since the inscribed angle (angle,ACB) and the central angle (angle,AOB) intercept the same arc (overparen{AB} ), we know from Theorem 2.4 that (angle,ACB = frac{1}{2},angle,AOB ). Hence, (angle,ACB = angle,AOD ). So since (C =angle,ACB ), we have

[nonumber
sin;C ~=~ sin;angle,AOD ~=~ frac{AD}{OA} ~=~ frac{frac{c}{2}}{R} ~=~ frac{c}{2R}
quadRightarrowquad 2,R ~=~ frac{c}{sin;C} ~,
]

so by the Law of Sines the result follows if (O) is inside or outside (triangle,ABC ).

Now suppose that (O) is on (triangle,ABC ), say, on the side (overline{AB} ), as in Figure 2.5.2(c). Then (overline{AB}) is a diameter of the circle, so (C = 90^circ) by Thales' Theorem. Hence, (sin;C = 1 ), and so (2,R = AB = c = frac{c}{1} = frac{c}{sin;C}; ), and the result again follows by the Law of Sines. QED

Solution:

We know that (triangle,ABC) is a right triangle. So as we see from Figure 2.5.3, (sin;A = 3/5 ). Thus,

[ 2,R ~=~ frac{a}{sin;A} ~=~ frac{3}{frac{3}{5}} ~=~ 5 quadRightarrowquad boxed{R ~=~ 2.5} ~.nonumber ]

Note that since (R =2.5 ), the diameter of the circle is (5 ), which is the same as (AB ). Thus, (overline{AB}) must be a diameter of the circle, and so the center (O) of the circle is the midpoint of (overline{AB} ).

Corollary 2.6

For any right triangle, the hypotenuse is a diameter of the circumscribed circle, i.e. the center of the circle is the midpoint of the hypotenuse.

For the right triangle in the above example, the circumscribed circle is simple to draw; its center can be found by measuring a distance of (2.5) units from (A) along (overline{AB} ).

We need a different procedure for acute and obtuse triangles, since for an acute triangle the center of the circumscribed circle will be inside the triangle, and it will be outside for an obtuse triangle. Notice from the proof of Theorem 2.5 that the center (O) was on the perpendicular bisector of one of the sides ((overline{AB})). Similar arguments for the other sides would show that (O) is on the perpendicular bisectors for those sides:

Corollary 2.7

For any triangle, the center of its circumscribed circle is the intersection of the perpendicular bisectors of the sides.

Example 2.18

Find the radius (R) of the circumscribed circle for the triangle (triangle,ABC) from Example 2.6 in Section 2.2: (a = 2 ), (b = 3 ), and (c = 4 ). Then draw the triangle and the circle.

Solution:

In Example 2.6 we found (A=28.9^circ ), so (2,R = frac{a}{sin;A} = frac{2}{sin;28.9^circ} = 4.14 ), so (boxed{R = 2.07}; ).
In Figure 2.5.5(a) we show how to draw (triangle,ABC): use a ruler to draw the longest side (overline{AB}) of length (c=4 ), then use a compass to draw arcs of radius (3) and (2) centered at (A) and (B ), respectively. The intersection of the arcs is the vertex (C ).

In Figure 2.5.5(b) we show how to draw the circumscribed circle: draw the perpendicular bisectors of (overline{AB}) and (overline{AC}); their intersection is the center (O) of the circle. Use a compass to draw the circle centered at (O) which passes through (A ).

Theorem 2.5 can be used to derive another formula for the area of a triangle:

Theorem 2.8

For a triangle (triangle,ABC ), let (K) be its area and let (R) be the radius of its circumscribed circle. Then

[K ~=~ frac{abc}{4,R} quad ( text{and hence }; R ~=~ frac{abc}{4,K} ~) ~. label{2.36}]

To prove this, note that by Theorem 2.5 we have

[nonumber
2,R ~=~ frac{a}{sin;A} ~=~ frac{b}{sin;B} ~=~ frac{c}{sin;C} quadRightarrowquad
sin;A ~=~ frac{a}{2,R} ~,~~ sin;B ~=~ frac{b}{2,R} ~,~~ sin;C ~=~ frac{c}{2,R} ~.
]

Substitute those expressions into Equation 2.26 from Section 2.4 for the area (K):

[nonumber
K ~=~ frac{a^2 ;sin;B ;sin;C}{2;sin;A} ~=~
frac{a^2 ;cdot; frac{b}{2,R} ;cdot; frac{c}{2,R}}{2;cdot; frac{a}{2,R}}
~=~ frac{abc}{4,R} qquad textbf{QED}
]

Combining Theorem 2.8 with Heron's formula for the area of a triangle, we get:

Corollary 2.9

For a triangle (triangle,ABC ), let (s = frac{1}{2}(a+b+c) ). Then the radius (R) of its circumscribed circle is

[R ~=~ frac{abc}{4,sqrt{s,(s-a),(s-b),(s-c)}} ~~.label{2.37}]

Circumscribed Rectangle, Bounding Box. (Coordinate Geometry) - Math ...

Let (r) be the radius of the inscribed circle, and let (D ), (E ), and (F) be the points on (overline{AB} ), (overline{BC} ), and (overline{AC} ), respectively, at which the circle is tangent. Then (overline{OD} perp overline{AB} ), (overline{OE} perp overline{BC} ), and (overline{OF} perp overline{AC} ). Thus, (triangle,OAD) and (triangle,OAF) are equivalent triangles, since they are right triangles with the same hypotenuse (overline{OA}) and with corresponding legs (overline{OD}) and (overline{OF}) of the same length (r ). Hence, (angle,OAD =angle,OAF ), which means that (overline{OA}) bisects the angle (A ). Similarly, (overline{OB}) bisects (B) and (overline{OC}) bisects (C ). We have thus shown:

For any triangle, the center of its inscribed circle is the intersection of the bisectors of the angles.

We will use Figure 2.5.6 to find the radius (r) of the inscribed circle. Since (overline{OA}) bisects (A ), we see that (tan;frac{1}{2}A = frac{r}{AD} ), and so (r = AD ,cdot, tan;frac{1}{2}A ). Now, (triangle,OAD) and (triangle,OAF) are equivalent triangles, so (AD =AF ). Similarly, (DB = EB) and (FC = CE ). Thus, if we let (s=frac{1}{2}(a+b+c) ), we see that
[nonumberbegin{align*}
2,s ~&=~ a ~+~ b ~+~ c ~=~ (AD + DB ) ~+~ (CE + EB) ~+~ (AF + FC) nonumber
&=~ AD ~+~ EB ~+~ CE ~+~ EB ~+~ AD ~+~ CE ~=~ 2,(AD + EB + CE) nonumber
s ~&=~ AD ~+~ EB ~+~ CE ~=~ AD ~+~ a nonumber
AD ~&=~ s - a ~.
end{align*}]
Hence, (r = (s-a),tan;frac{1}{2}A ). Similar arguments for the angles (B) and (C) give us:

Theorem 2.10

For any triangle (triangle,ABC ), let (s = frac{1}{2}(a+b+c) ). Then the radius (r) of its inscribed circle is

[ r ~=~ (s-a),tan;tfrac{1}{2}A ~=~ (s-b),tan;tfrac{1}{2}B ~=~
(s-c),tan;tfrac{1}{2}C ~.label{2.38}]

We also see from Figure 2.5.6 that the area of the triangle (triangle,AOB) is

[nonumber
text{Area}(triangle,AOB) ~=~ tfrac{1}{2},text{base} times text{height} ~=~
tfrac{1}{2},c,r ~.
]

Similarly, (text{Area}(triangle,BOC) = frac{1}{2},a,r) and (text{Area}(triangle,AOC) = frac{1}{2},b,r ). Thus, the area (K) of (triangle,ABC) is

[nonumber begin{align*}
K ~&=~ text{Area}(triangle,AOB) ~+~text{Area}(triangle,BOC) ~+~ text{Area}(triangle,AOC)
~=~ tfrac{1}{2},c,r ~+~ tfrac{1}{2},a,r ~+~ tfrac{1}{2},b,r nonumber
&=~ tfrac{1}{2},(a+b+c),r ~=~ sr ~,~text{so by Heron's formula we get} nonumber
r ~&=~ frac{K}{s} ~=~ frac{sqrt{s,(s-a),(s-b),(s-c)}}{s} ~=~
sqrt{frac{s,(s-a),(s-b),(s-c)}{s^2}} ~=~ sqrt{frac{(s-a),(s-b),(s-c)}{s}} ~~.
end{align*}]

We have thus proved the following theorem:

Theorem 2.11

For any triangle (triangle,ABC ), let (s = frac{1}{2}(a+b+c) ). Then the radius (r) of its inscribed circle is

[label{2.39}r ~=~ frac{K}{s} ~=~ sqrt{frac{(s-a),(s-b),(s-c)}{s}} ~~.]

Recall from geometry how to bisect an angle: use a compass centered at the vertex to draw an arc that intersects the sides of the angle at two points. At those two points use a compass to draw an arc with the same radius, large enough so that the two arcs intersect at a point, as in Figure 2.5.7. The line through that point and the vertex is the bisector of the angle. For the inscribed circle of a triangle, you need only two angle bisectors; their intersection will be the center of the circle.

Example 2.19

Find the radius (r) of the inscribed circle for the triangle (triangle,ABC) from Example 2.6 in Section 2.2: (a = 2 ), (b = 3 ), and (c = 4 ). Draw the circle.

Figure 2.5.8

Solution:

Using Theorem 2.11 with (s = frac{1}{2}(a+b+c) =frac{1}{2}(2+3+4) = frac{9}{2} ), we have

[ r ~=~ sqrt{frac{(s-a),(s-b),(s-c)}{s}} ~=~
sqrt{frac{left(frac{9}{2}-2right),left(frac{9}{2}-3right),left(frac{9}{2}-
4right)}{frac{9}{2}}} ~=~ sqrt{frac{5}{12}}~.nonumber ]

Figure 2.5.8 shows how to draw the inscribed circle: draw the bisectors of (A) and (B ), then at their intersection use a compass to draw a circle of radius (r = sqrt{5/12} approx 0.645 ).

Contributors and Attributions

Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2.

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We obtained an unique take a look at the pitch deck UK fintech Payhawk utilized to increase $20 million from Klarna backer QED

We obtained an unique take a look at the pitch deck UK fintech Payhawk utilized to increase $20 million from Klarna backer QED

Summary List Placement London fintech Payhawk has actually increased
$20 million in a round led by United States fund QED Investors. The settlements and also expenditure monitoring system asserts to have actually expanded 10 times over in 2015 as a result of a rise sought after from the pandemic. Payhawk, which was established in 2018, declares it boosts the performance of a business’s…

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Payhawk raises $20M to unify corporate cards, payments and expenses – | #Tech

Payhawk raises $20M to unify corporate cards, payments and expenses – | #Tech

Fintech startup Payhawk has raised a $20 million funding round. QED Investors is leading the round with existing investor Earlybird Digital East also participating. Payhawk is building a unified system to manage all the money that is going in and out.
Essentially, companies switching to Payhawk can replace several services they already use and that didn’t interact well with each other. Payhawk…

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Payhawk raises $20M to unify corporate cards, payments and expenses

Fintech startup Payhawk has raised a $20 million funding round. QED Investors is leading the round with existing investor Earlybird Digital East also participating. Payhawk is building a unified system to manage all the money that is going in and out.

Essentially, companies switching to Payhawk can replace several services they already use and that didn’t interact well with each other. Payhawk lets you issue corporate cards for your employees, manage invoices and track payments from a single interface.

After signing up, customers get their own banking details with a dedicated IBAN. You can connect with your existing bank account, load funds to your Payhawk account and start using it in multiple ways.

Compared to other companies working on similar products, Payhawk gives each customer their own IBAN, which means they can receive third-party payments.

One of the key features of Payhawk is that customers can issue virtual and physical cards for employees with different rules. You can set up a team budget, configure an approval workflow for large transactions and let Payhawk handle receipt collection from those card transactions.

You can upload invoices to manage them through Payhawk. The startup tries to automatically extract data from those invoices for easier reconciliation. Payhawk also lets you reimburse employees. The service acts as a single source of truth for your company’s spending. Finally, you can connect Payhawk with your existing ERP system.

As a software-as-a-service solution, you pay a monthly subscription fee that will vary depending on optional features and the number of active cards. Clients include LuxAir, Lotto24, Viking Life, ATU, Gtmhub, MacPaw and By Miles. Overall, the startup has 200 clients.

The company has been growing nicely as revenue doubled in Q1 2021. It currently accepts clients in the European Union and the U.K. but it already plans to expand beyond those markets. Up next, Payhawk plans to launch credit cards, more currencies and tighter integration with corporate bank accounts.

Fintechs could see $100 billion of liquidity in 2021

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Payhawk raises $20 million to unify corporate cards, payments and expenses

Payhawk raises $20 million to unify corporate cards, payments and expenses

Fintech startup Payhawk has raised a $20 million funding round. QED Investors is leading the round with existing investor Earlybird Digital East also participating. Payhawk is building a unified system to manage all the money that is going in and out.
Essentially, companies switching to Payhawk can replace several services that they already use and that didn’t interact well with each other.…

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DelveInsight has launched new reports: Bile Duct Neoplasm Market, Dermatomycoses Market, Familial Primary Pulmonary Hypertension Market and Intravenous Immunoglobulin Market

Bile Duct Neoplasm Market

DelveInsight's "Bile Duct Neoplasms Market Insights, Epidemiology, and Market Forecast-2030" report delivers an in-depth understanding of the Bile Duct Neoplasms , historical and forecasted epidemiology as well as the Bile Duct Neoplasms market trends in the United States, EU5 (Germany, Spain, Italy, France, and United Kingdom) and Japan.

Bile Duct Neoplasm Companies:

· Agios Pharmaceuticals

· AstraZeneca

· Merck/EMD Serono

· QED Therapeutics

· Nucana

· Lexicon Pharmaceutical

· And Many Others

View Report: https://www.delveinsight.com/report-store/bile-duct-neoplasms-market

Dermatomycoses Market

DelveInsight's "Dermatomycoses Market Insights, Epidemiology, and Market Forecast-2030" report delivers an in-depth understanding of the Dermatomycoses , historical and forecasted epidemiology as well as the Dermatomycoses market trends in the United States, EU5 (Germany, Spain, Italy, France, and United Kingdom) and Japan.

Dermatomycoses Companies:

· Anacor Pharmaceuticals Inc

· Biolab Farmaceutica Ltda

· Blueberry Therapeutics Ltd

· Daewoong Pharmaceutical Co Ltd

· Dermala Inc

· Helix BioMedix Inc

· Novan Inc

· Sol-Gel Technologies Ltd

· TGV-Laboratories

· Viamet Pharmaceuticals Inc

· Vyome Biosciences Pvt Ltd

View Report: https://www.delveinsight.com/report-store/dermatomycoses-market

Familial Primary Pulmonary Hypertension Market

DelveInsight's "Familial Primary Pulmonary Hypertension Market Insights, Epidemiology, and Market Forecast-2030" report delivers an in-depth understanding of the Familial Primary Pulmonary Hypertension , historical and forecasted epidemiology as well as the Familial Primary Pulmonary Hypertension market trends in the United States, EU5 (Germany, Spain, Italy, France, and United Kingdom) and Japan.

Familial primary pulmonary hypertension is a rare autosomal dominant disorder that has reduced penetrance and that has been mapped to a 3-cM region on chromosome 2q33 (locus PPH1). The phenotype is characterized by monoclonal plexiform lesions of proliferating endothelial cells in pulmonary arterioles. These lesions lead to elevated pulmonary-artery pressures, right-ventricular failure, and death. Although primary pulmonary hypertension is rare, cases secondary to known etiologies are more common and include those associated with the appetite-suppressant drugs, including phentermine-fenfluramine.

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Intravenous Immunoglobulin Market

Intravenous immunoglobulin (also referred to as IVIG) is a product made up of antibodies that can be given intravenously. According to the American College of Rheumatology, it is prepared from a pool of immunoglobulins (antibodies) from the plasma of thousands of healthy donors.

Immunoglobulins are made by the immune system of healthy people to fight infections. While IVIG is derived from plasma (a blood product), it is so purified that the chances of contracting a blood-borne infection are extremely low.

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· Biotest

· Momenta Pharmaceuticals

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· And Many Others

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Glass Balustrade Lamination Market - Global Industry Report 2030

Groundwater Remediation Technologies Market: Introduction

Groundwater remediation is the process of treating the polluted groundwater by eliminating harmful pollutants or converting them into harmless products. Groundwater is the water present in the earth’s surface that saturates the pore space in the subsurface. Usually, about 20% to 40% of drinking water is drawn from digging wells and boreholes in the world.

Groundwater is one of the primary sources of drinking water. It is also used for industrial and agricultural usage. However, the quality of groundwater has been deteriorating due to unorganized waste disposal practices and accidental spillage of hazardous chemicals. Thus, it becomes critical to prevent groundwater pollution. Contaminated groundwater must be remediated in order to protect public health and the surrounding environment.

Contaminants or pollutants found in groundwater cover a broad range of physical, organic chemical, inorganic chemical, bacteriological, and radioactive parameters. These parameters need to be removed from groundwater by applying various remediation techniques in order to bring the groundwater to a standard condition for intended use.

Contaminants encountered in groundwater include organic compounds, heavy metals, and radionuclides. These contaminants are generated from a number of sites such as Underground Storage Tanks (UGTs) and various civilian federal agency sites.

Remediation of groundwater pollution needs to be carried out depending on specific water quality, environmental conditions, and the amount of mix of pollutants present in it

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Key Drivers of Global Groundwater Remediation Technologies Market

Groundwater continues to serve as the major reliable source of water for a variety of purposes, including industrial and domestic usage, and irrigation. Remediation of groundwater is necessary to remove contaminants and components that can harm human health and the surrounding environment.

Groundwater contamination by various industries remains a widespread concern. The prevalence of contaminants at hazardous waste sites is well documented. These contaminates can contaminate millions of liters of groundwater if they are not eliminated or sequestered. This is anticipated to drive the groundwater remediation technologies market during the forecast period.

The usage of agricultural chemicals has been low in many developing countries around the world compared to the levels in industrialized countries. Irrigation is carried out extensively, particularly in countries such as India and China. Concerns over groundwater pollution from agricultural chemicals were raised in India more than two decades ago. This is estimated to boost the groundwater remediation technologies market during the forecast period.

Government regulatory bodies across the world are seeking to introduce innovative technologies, conduct surveys on groundwater contamination, and classify the groundwater into categories according to its quality. Additionally, corresponding measures and risk evaluation systems are being adopted to curb groundwater pollution. This is projected to propel the groundwater remediation technologies market during the forecast period.

Efforts to encourage the formation of groundwater user associations are in the nascent stage in developing countries. The results of irrigation management transfer and water user associations are now being documented. This is likely to augment the groundwater remediation technologies market in the near future.

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Key Developments

In April 2020, the National Ground Water Association in the U.S. published an article on the potential risk of COVID-19 in contaminating private water well systems. The NGWA Director of Science and Technology reported that private water wells were at low risk of transmitting COVID-19.

In March 2020, the U.S. Department of Energy announced that environmental management workers were making progress in upgrading aging radiological wastewater treatment infrastructure essential to Oak Ridge National Laboratory (ORNL) operations. They were installing a new zeolite treatment system, which is designed to remove cesium and strontium from wastewater.

In February 2020, The U.S. Environmental Protection Agency announced plans to submit an information collection request for the 2020 Drinking Water Infrastructure Needs Survey and Assessment (DWINSA) to the Office of Management and Budget (OMB) for review and approval in accordance with the Paperwork Reduction Act

Asia Pacific to Hold Major Share of Global Groundwater Remediation Technologies Market

Based on region, the global groundwater remediation technologies market can be classified into North America, Asia Pacific, Europe, Latin America, and Middle East & Africa

Asia Pacific is expected to dominate the global market in the near future, as rapid urbanization and industrialization in major economies such as China, South Korea, and India have resulted in groundwater pollution

Initiatives toward remediation of groundwater from renewable energy by governments in countries of Asia Pacific are expected to positively impact the global groundwater remediation technologies market during the forecast period

In Europe, The Groundwater Directive (2006/118/EC) has established a regime that sets groundwater quality standards and introduces measures to prevent or limit inputs of pollutants into groundwater. The directive establishes quality criteria that takes into account local characteristics and allows for further improvements to be made based on monitoring data and new scientific knowledge. Thus, the directive represents a proportionate and scientifically sound response to the requirements of the Water Framework Directive (WFD) as it relates to assessment on the chemical status of groundwater and the identification and reversal of significant and sustained upward trends in pollutant concentrations.

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Key Players Operating in Global Groundwater Remediation Technologies Market:

Prominent players operating in the global groundwater remediation technologies market include:

QED Environmental Systems

Itasca International Inc.

ESCO International (EI)

Dürr AG

Etaniv B.Y ltd

Lenntech B.V.

Kurita Water Industries Ltd

Global Groundwater Remediation Technologies Market, Research Scope

Global Groundwater Remediation Technologies Market, by Source

Industrial Waste

Agricultural Waste

Municipal Waste

Radioactive Waste

Others

Global Groundwater Remediation Technologies Market, by Type

Pump & Treat

In-situ Air Sparging

In-Situ Flushing

Permeable Reactive Barrier

Others

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