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3 years ago

The circumference of a circle is 9 pi.What is the area,in square inches,of a circle?Express your answer in terms of pi.

Mathematics

1 answer:

Alla [95]3 years ago

5 0

Answer:

20.25π

Step-by-step explanation:

The circumference (C) of a circle is calculated using the formula

C = 2πr ← r is the radius

given C = 9π, then

2πr = 9π ( divide both sides by 2π )

r = $\frac{9\pi }{2\pi }$ ( cancel the π on numerator/denominator )

= 4.5

The area (A) of a circle is calculated using the formula

A = πr² = π × 4.5² = 20.25π

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Evaluate: 2·5/4-2 A) 5/8 B) 8/5 C) 40 D) 160

allochka39001 [22]

Answer: 1/2

Multiply

2×5=10

10/4-2 is your new problem

Simplify

10/4 = 5/2

5/2-2 is your new problem

Subtract

5/2-2=1/2

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2 years ago

A triangle has sides with lengths of 2x - 7, 5x - 3, and 2x - 2. What is the perimeter

alex41 [277]

P=2x-7 + 5x-3 +2x-2
P= 9x -12

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2 years ago

A. 0 B. 1 C. 2 D. 3

zloy xaker [14]

-19 to -15 would be a positive 4 difference

19 to 15 = 4 this would be a -4 difference

3 to 7 is a positive 4 difference

-3 to 7 is a difference of 10

so there is 1 with a -4 difference
answer is B.

6 0

2 years ago

Delicious77 [7]

You can solve for the velocity and position functions by integrating using the fundamental theorem of calculus:

a(t) = 40 ft/s²

v(t) = v (0) + ∫₀ᵗ a(u) du

v(t) = -20 ft/s + ∫₀ᵗ (40 ft/s²) du

v(t) = -20 ft/s + (40 ft/s²) t

s(t) = s (0) + ∫₀ᵗ v(u) du

s(t) = 10 ft + ∫₀ᵗ (-20 ft/s + (40 ft/s²) u ) du

s(t) = 10 ft + (-20 ft/s) t + 1/2 (40 ft/s²) t ²

s(t) = 10 ft - (20 ft/s) t + (20 ft/s²) t ²

6 0

2 years ago

1. y = x2 + 8x + 15 Find the zeros of the function by rewriting the function in intercept form

lubasha [3.4K]

The zeros of given function $y=x^{2}+8 x+15$ is – 5 and – 3

Solution:

$\text { Given, equation is } y=x^{2}+8 x+15$

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as to obtain zeros of function

We know that, intercept form of above equation is $x^{2}+8 x+15=0$

$\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3$

$\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}$

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0

3 years ago