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vesna_86 [32]
3 years ago
7

George plans to cover his circular pool for the upcoming winter season. The pool has a diameter of 20 feet and the cover extends

12 inches beyond the edge of the pool. A rope runs along the edge of the cover to secure it in place.
a.
What is the area of the pool cover?
b. What is the length of the rope?

You must show all of your work to receive credit
Mathematics
1 answer:
Verizon [17]3 years ago
8 0

Answer:

Area of Pool cover: 110.25π

Length of rope: 21π

Step-by-step explanation:

To find the area of a circle the formula is πr^2.

1. take 10.5, which is half of the diameter and multiply it by itself then multiply by pi(π). Answer is 110.25π

To find the length of the rope(circumference) the formula is 2πr.

1. Take r, which we established as 10.5 then multiply it by 2, which is 21. Then multiply by π. Answer is 21π

____________________________________________________________

If ur teacher wants you to use 3.14 as pi then these are your answers:

Area of Pool cover: 346.185, round up to 346.19 if needed

Length of rope: 65.94

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Answer:

\frac{3}{8} or\ 0.375\ km

Step-by-step explanation:

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First, lets find how many 10 minutes in one hours or convert minutes in hours.

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Now, finding kilometres travelled in 1 hour.

Distance travelled= \frac{1}{16} \times \frac{60}{10} = \frac{1}{16} \times 6

Distance travelled = \frac{3}{8}

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Find dy/dx if y =x^3+5x+2/x²-1
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<u>Differentiate using the Quotient Rule</u> –

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According to the given question, we have –

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Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

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En un 9no grado, la asignatura de matemáticas la han aprobado el 62.5% de las alumnas y el 80% de los alumnos, mientras que la a
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Answer:

In the classroom there are 16 girls and 20 boys

Step-by-step explanation:

The question in English is

In the ninth grade, 62.5% of girls and 80% of boys passed the mathematics course, while 87.5% of girls and 60% of boys passed the history course. Calculate the number of students in the classroom, if the total number of passes is 26 in both subjects

Let

x ----> total girls in the classroom

y ---> total boys in the classroom

we know that

mathematics course

0.625x+0.80y=26 -----> equation A

history course

0.875x+0.60y=26 ----> equation B

Solve the system of equations by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

The solution is the point (16,20)

see the attached figure

therefore

In the classroom there are 16 girls and 20 boys

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