What is the lateral surface area on a square pyramid with base edges of 10 and the height of 12

At a local department store, skirts typically cost $15. However, due to a special, the

Strike441 [17]

Answer: 20%

Step-by-step explanation:

(part / whole) * 100

(3 / 15) * 100 = 20%

4 0

3 years ago

Which is a true statement about 7 and 14?

Ksju [112]

7 and 14 can be divided in to two

7 0

3 years ago

A band expects to put 12 songs on their next CD. The band writes and records 25% more songs than they expect to put on the CD.

sweet-ann [11.9K]

12 songs. 25% of 12 is 3, so you add 3 to 12 and get 15. You then find that 20% of 15 is 3 so you subtract 3 from 15 and get 12 again.

7 0

3 years ago

Read 2 more answers

Consider the following sequence

wlad13 [49]

Answer:

a₇ = 2.375

Step-by-step explanation:

There is a common ratio r between consecutive terms, that is

r = $\frac{-76}{152}$ = $\frac{38}{-76}$ = $\frac{-19}{38}$ = - $\frac{1}{2}$

This indicates the sequence is geometric with nth term

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio

Here a₁ = 152 and r = - $\frac{1}{2}$ , then

$a_{n}$ = 152 $(-\frac{1}{2}) ^{n-1}$ , so

a₇ = 152 $(-\frac{1}{2}) ^{6}$ = 152 × $\frac{1}{64}$ = 2.375

3 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

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Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago