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3 years ago

In a department store, a $190 phone is marked, "Save 15%."

Mathematics

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

a- $28.50

b- $161.50

Step-by-step explanation:

multiply 190 and 0.15 and you get 28.5 subtract that from 190 and you get 161.5

Hope this helps!

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Adam traveled out of town for a regional basketball tournament. He drove at a steady speed of 72.4 miles per hour for 4.62 hours

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Answer:

Step-by-step explanation:

334.488 miles

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3 years ago

A rectangular park is 5/6 miles wide and 1 5/7 miles long.

vladimir2022 [97]

Area = Length × width

Area = 5/6 × 15/7 = 75/42 mi²

Area = $1 \displaystyle\frac{33}{42}$

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3 years ago

Find the value of z such that 0.11 of the area lies to the right of z. Round your answer to two decimal places.

Art [367]

Answer:

1.23?

Step-by-step explanation:

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3 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

1 year ago

The picture is above I’ll mark as brainliest.

EastWind [94]

Answer:

<h2>7 inches</h2>

Step-by-step explanation:

see attached photo

7 0

3 years ago

Read 2 more answers