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3 years ago

When factored completely the expression p^4-81 is equivalent to

1 answer:

7 0

$\displaystyle\\
p^4 - 81 =\\\\
 =p^4-3^4 =\\\\
= \Big(p^2\Big)^2 - \Big(3^2\Big)^2 =\\\\
=\Big(p^2+3^2\Big)\Big(p^2-3^2\Big)=\\\\
=\boxed{\bf \Big(p^2+3^2\Big)\Big(p+3\Big)\Big(p-3\Big)}$

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Simplify : 26-10/2x5+3

zvonat [6]

Answer:

16/13 is the answer.

26-10 is 16 and 2x5+3 is 13

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3 years ago

Read 2 more answers

Elisa is nearing graduation. She just met with her counselor and he showed her this information.

Arada [10]

Answer: $14,000

Step-by-step explanation:

Given the following :

Room & Board = $11,500

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Grant & scholarship = $9500

Tuition fee = $9500

If she pays balance with a student loan, how much will she need to borrow:

Elisha's bill:

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($9500 + $11500) = $21,000

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Therefore,

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3 years ago

Solve proportion round it to the nearest hundredth

Sunny_sXe [5.5K]

Answer:

7.7 = 2.3

3.6 b

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b = 1.075324675

b = 1.08

A) If the last digit in the fractional part of 1.080 is less than 5, then simply remove the last the digit of the fractional part. With 1.080, rule A applies and 1.08 rounded to the nearest hundredth is:

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3 years ago

Given an=−6+(n−1)(23) a n = - 6 + ( n - 1 ) ( 2 3 ) what would the constant rate of change be?

KiRa [710]

23 is the constant rate of change..............

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3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

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2 years ago