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Sophie [7]
2 years ago
6

Ella earned a 16/20 on her quiz. What percent is this​

Mathematics
1 answer:
kari74 [83]2 years ago
7 0

Answer:

80%

Step-by-step explanation:

1. 16/20 = 0.8

2. 0.8 * 100 = 80%

Therefore the answer is 80%, hope this helps!

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Write the domain of the function using interval notation
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(-♾, 5) (5, ♾)

Step-by-step explanation:

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Abbey purchased a poster that had a height of 2 feet and a width of 0.7 foot.What is the area of Abbey's poster?
Eduardwww [97]

Answer:

1.4 feet

Step-by-step explanation:

To determine area we multiply height X width

2 X .7 = 1.4

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Find the first and second derivatives of the function y = sin(x^2)
exis [7]
Y = sin(x^2)

Use the chain rule.

We want dy/dx.

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dy/dx = 2xcos(x^2)

Did you follow?
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How do i do parallel lines and transversals
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A transversal is a line that intersects two or more other lines. When it intersects parallel lines, many angles are congruent. Parallel lines<span> are two </span>lines<span> that are always the same distance apart and never touch.</span>
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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}

8 0
3 years ago
Read 2 more answers
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