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3 years ago

11

Whats is the product (6r-1)(-8-3)

1 answer:

iren [92.7K]3 years ago

7 0

Go to Mathaway! Great math solver for these problems!! :)

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How to write 0.624 in expanded form

GalinKa [24]

Answer:

0.6 + 0.02 + 0.004

Step-by-step explanation:

0.624=0.6 + 0.02 + 0.004

6 0

3 years ago

Aiden buys two bags of chips at $1.09 each and a 20 ounce soda for $1.20 at the gas station.He knows that the state tax is 5.5.

Natalka [10]

Aiden will receive $1.43 from a $5 bill.

If the state tax is 5.5%, you can use the equation below to solve for the total cost, <em>t</em>.

<em>t</em> = 1.055 (2 × 1.09 + 1.20)
<em>t</em> = 1.055 (2.18 + 1.20)
<em /><em>t </em>= 1.055 (3.38)
<em>t </em>= 3.5659

You can round 3.5659 to $3.57. Finally, you subtract 3.57 from 5 to find the amount of change Aiden will receive, <em>c</em><em />.

<em>c </em>= 5 - 3.57
<em>c </em>= 1.43

7 0

3 years ago

................. is a factor of every number

poizon [28]

Answer:

The answer is 1.

Step-by-step explanation:

One is the divisor of every integer. Or, every whole number is the product of 1 and itself

6 0

2 years ago

Please help 20 points to who ever can

mixer [17]

Answer:

how is your mom doing tell her last night was crazy :D

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

6 0

3 years ago