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sweet [91]
2 years ago
10

10 The diagram shows a garden in the shape of a rectangle.

Mathematics
1 answer:
harkovskaia [24]2 years ago
3 0
20 + 8x = 32
- 20
8x = 12
--      --
8       8

x = 3/2= 1 1/2
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2x^2+6x+3=0<br> Solve for quadratic formula
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Answer:

x= (-3 +- sqrt3)/2

Step-by-step explanation:

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What is the best approximation for the circumference of a circle with a diameter of 300 feet
garri49 [273]

Answer:

The answer is C=942.48

Step-by-step explanation:

You would figure this out by using the formula C=2πr.

So let's break this down.

Circumference (C) equals (=)  pi (π) times 2 times the radius of the circle.

Since the problem gave you the diameter which is the length from one side of the circle to the other, you have to divide the diameter in half. Which would give you a radius of 150.

Taking the 150 you would multiply 10*2*π.

Which would give you....(drumrollllll)

C=942.48

I hope this helps!!

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Tony rounded each of the numbers 1, 143 and 1, 149
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last one.................

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3 years ago
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I need help on this question?​
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2 years ago
Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.
Lilit [14]

For starters,

\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k

Consider the nth partial sum, denoted by S_n:

S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n

Multiply both sides by \frac34:

\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}

Subtract S_n from this:

\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)

Solve for S_n:

-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)

S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)

Now as n\to\infty, the exponential term will converge to 0, since r^n\to0 if 0. This leaves us with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}

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3 years ago
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