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oksian1 [2.3K]
3 years ago
6

Pease answer number 9 for me.

Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

a. 33 degrees

b. 90 degrees

c. 57 degrees

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Is 80 homework and 74 exit tickets equivalent?
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No. If one student each handed in their homework but six did not hand in an exit ticket.
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What is the distance between the points (−5, −9)(−5, −9) and (−5, 13)(−5, 13) ?
krok68 [10]
Distance formula : d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(-5,-9)...x1 = -5, and y1 = -9
(-5,13)...x2 = -5 and y2 = 13
now we sub and solve
d = sqrt (-5 - (-5)^2 + (13 - (-9)^2
d = sqrt (-5 + 5)^2 + (13 + 9)^2
d = sqrt (0^2 + 22^2)
d = sqrt (484)
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Tony earns $100 for walking 5 dogs. how much would he earn for each dog​
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Solve for z<br> -p(d+z) = -2z+59−p(d+z)=−2z+59
alexira [117]

-p(d+z)=-2z+59\qquad\text{use distributive property}\\\\-pd-pz=-2z+59\qquad\text{add pd to both sides}\\\\-pz=-2z+pd+59\qquad\text{add 2z to both sides}\\\\2z-pz=pd+59\\\\z(2-p)=pd+59\qquad\text{divide both sides by}\ (2-p)\neq0\\\\\boxed{z=\dfrac{pd+59}{2-p}}

6 0
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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
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