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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Mathematics

1 answer:

Rashid [163]3 years ago

3 0

$x = \frac{ -2 \pm \: \sqrt{ {2}^{2} - 4 \times 1 \times 10 } }{2} \\ \\ x = \frac{ - 2 \pm 6i}{2} \\ \\ \boxed{ \red{ \huge{x = - 1 \pm \: 3i}}}$

$\boxed{ \red{ \huge{answer = d}}}$

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The figure shows the layout of a symmetrical pool in a water park. What is the area of this pool rounded to the tens place? Use

k0ka [10]

Answer:

$2489ft^{2}$

Step-by-step explanation:

The pool are is divided into 4 separated shapes: 2 circular sections and 2 isosceles triangles. Basically, to calculate the whole area, we need to find the area of each section. Due to its symmetry, both triangles are equal, and both circular sections are also the same, so it would be enough to calculate 1 circular section and 1 triangle, then multiply it by 2.

<h3>Area of each triangle:</h3>

From the figure, we know that <em>b = 20ft </em>and <em>h = 25ft. </em>So, the area would be:

$A_{t}=\frac{b.h}{2}=\frac{(20ft)(25ft)}{2}=250ft^{2}$

<h3>Area of each circular section:</h3>

From the figure, we know that $\alpha =2.21 radians$ and the radius is $R=30ft$ . So, the are would be calculated with this formula:

$A_{cs}=\frac{\pi R^{2}\alpha}{360\°}$

Replacing all values:

$A_{cs}=\frac{(3.14)(30ft)^{2}(2.21radians)}{6.28radians}$

Remember that $360\°=6.28radians$

Therefore, $A_{cs}=994.5ft^{2}$

Now, the total are of the figure is:

$A_{total}=2A_{t}+2A{cs}=2(250ft^{2} )+2(994.5ft^{2})\\A_{total}=500ft^{2} + 1989ft^{2}=2489ft^{2}$

Therefore the area of the symmetrical pool is $2489ft^{2}$

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