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2 years ago

HELP ASAP!! it’s a test :( thank you if you helped

Mathematics

1 answer:

olchik [2.2K]2 years ago

5 0

Answer:

its I, the last graph

Step-by-step explanation:

if you look at the first number of each point its plus 3 and its the same for the last numbers.

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What is y - 8 = -1/2 (x + 4) in standard form (8th grade math)

Scilla [17]

This equation is in point-slope form. To change it to standard form, you must distribute -1/2 to (x+4) to get -1/2x-2. Your equation should now be y-8 = -1/2x-2. You should then add 1/2x to both sides (to cancel out -1/2x on the right side). Now, you should have y+1/2x-8 = -2. Finally, add 8 to both sides (to cancel out -8 on the left side) and flip the x and the y to get a final answer of 1/2x+y = 6.

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3 years ago

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What is the equation of the horizontal asymptote? f(x)=4(52)x+7<br> y=?

natta225 [31]

ANSWER

y=7

EXPLANATION

The horizontal asymptote of an exponential function

$f(x)= a {(b)}^{x} + c$

is y=c.

The given exponential function is

$f(x)= 4 {(52)}^{x} + 7$

When we compare to

$f(x)= a {(b)}^{x} + c$

c=7, therefore the horizontal asymptote is y=7.

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2 years ago

compare the method of using expanded form and the method of using place value to multiply a decimal and a whole number

Zigmanuir [339]

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3 years ago

What is 3/14 divided by 1/2 (Reduce to lowest terms!)

cricket20 [7]

The answer to that question is 3/7

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2 years ago

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Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.

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3 years ago