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10 days ago

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Explaining How to Compare Water Levels Ericka decided to compare her observation to the average annual trend, which shows the wa

ter rising 1.8 mm/year. Remember, she used 6.2 years as her time period. Explain how she would calculate the difference between how much water levels rose on average and how much the water level fell in the part of the river she observed.

2 answers:

PIT_PIT [9.1K]10 days ago

7 0

To compute the average rise in water levels, she will need to multiply the annual increase of 1.8 by the duration of 6.2 years, resulting in 11.16. Next, to determine the difference between average water level rise and the observed decline of -13.64, she will subtract -13.64 from 11.16.

Inessa [9K]10 days ago

3 0

The explanation is as follows: a visual attachment including the solution is provided.

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The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

Svet_ta [9518]

Answer:

The shaped region accounts for 7/18 of the area of ACIG.

Step-by-step explanation:

Refer to the attached diagram for further clarity on the problem.

Step 1

Determine the length of one side of square ABED.

We know that

AB=BE=ED=AD

The area of a square can be calculated as

$A=b^{2}$

where b is the side length.

We have

$A=49\ units^2$

So we substitute

$49=b^{2}$

$b=7\ units$

Thus,

$AB=BE=ED=AD=7\ units$

Step 2

Calculate the area of ACIG.

The area of rectangle ACIG is determined by

$A=(AC)(AG)$

Substituting the given values yields

$A=(9)(10)=90\ units^2$

Step 3

Determine the area of the shaded rectangle DEHG.

The area of rectangle DEHG is given by

$A=(DE)(DG)$

We find $DE=7\ units$

$DG=AG-AD=9-7=2\ units$

and substitute $A=(7)(2)=14\ units^2$

Step 4

Calculate the area of shaded rectangle BCFE.

The area of rectangle BCFE equals

$A=(EF)(CF)$

We see that

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

and substitute

$A=(3)(7)=21\ units^2$

Step 5

Add the areas of the shaded regions together.

$14+21=35\ units^2$

Step 6

Divide the area of the shaded region by the area of ACIG.

$\frac{35}{90}$

Simplify this fraction by dividing both the numerator and denominator by 5.

$\frac{7}{18}$

Hence, the shaped region represents 7/18 of the area of ACIG.

5 0

1 month ago

out of fifty students,34 joined the journalism class and 25 joined robotics club.if 16 students joined both clubs,how many stude

AnnZ [9104]

Answer:

Step-by-step explanation:

Overall students=50

35 -> j.

25 -> r.

16-> j.+r.

35-16=19 students joined j.

25-16=9 joined r.

50-19+9

50-28=22

The answer is 22

It would be nice if you could give me the brainliest rating. Best of luck!

3 0

24 days ago

An upscale resort has built its circular swimming pool around a central area that contains a restaurant. The central area is a r

zzz [9093]

The diagram below illustrates the issue at hand.

Question 1:
The maximum area of the pool equals half the area of the circle.

To calculate the area of the circle: Area = πr², with r being half of the diameter.
Thus, Area of circle = π(60)² = 11309.73355 square feet.

Therefore, the area representing half the circle amounts to 11309.73355/2 = 5654.866... ≈ 5654.87 square feet (rounded to 2 decimal places).

Question b:

To find the pool's area, we take the circle's area and subtract the triangle's area.

The area of the circle is 11309.73 square feet.

For the triangle's area calculation: 1/2 × (60×103.92) = 3117.6 square feet.

The area of the pool thus operates as 11309.73 - 3117.6 = 7922.13 square feet.

Calculating the pool's volume: 7922.13 × 4 = 31688.52 cubic feet.

Note: Information related to the fish tank is unavailable, so the above calculation focuses solely on the entire pool's volume.

6 0

17 days ago

The graphs below have the same shape. What is the equation of the blue graph?

AnnZ [9104]

Answer: OPTION B

Step-by-step explanation:

The red graph depicts the fundamental form of a quadratic function (the most basic version), with its vertex located at the origin.

The function g(x) results from moving the parent function 2 units to the right and 1 unit upwards.

As a consequence, considering this, the transformation takes the following structure:

$g(x)=(x-h)^2+k$

The horizontal displacement is dictated by the value of h, while the vertical shift is determined by k.

<pThus, the resulting function is:

$g(x)=(x-2)^2+1$

5 0

1 month ago

Read 2 more answers

Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)

Leona [9271]

Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

$\Rightarrow x^2+y^2 = 16900$

$\Rightarrow y^2 = 16900-x^2$

$\Rightarrow y = \sqrt{16900-x^2}$

Since the point lies on the circle, its coordinates must be,

$(x,\sqrt{16900-x^2})$

The distance "d" from the point to (30,40) can be calculated as,

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}$

$=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}$

$=\sqrt{9400-60x-80\sqrt{16900-x^2}}$

Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when $9400-60x-80\sqrt{16900-x^2}$ is at its lowest value.

Let’s set up the equation,

$\Rightarrow f(x)=9400-60x-80\sqrt{16900-x^2}$

$\Rightarrow f'(x)=-60+80\dfrac{x}{\sqrt{16900-x^2}}$

$\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}$

We find the critical points,

$\Rightarrow f'(x)=0$

$\Rightarrow-60+80\dfrac{x}{\sqrt{16900-x^2}}=0$

$\Rightarrow 80\dfrac{x}{\sqrt{16900-x^2}}=60$

$\Rightarrow 80x=60\sqrt{16900-x^2}$

$\Rightarrow 80^2x^2=60^2(16900-x^2)$

$\Rightarrow 6400x^2=3600(16900-x^2)$

$\Rightarrow \dfrac{16}{9}x^2=16900-x^2$

$\Rightarrow \dfrac{25}{9}x^2=16900$

$\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78$

$\Rightarrow x=78$

Then,

$\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2$

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

5 0

1 month ago