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s344n2d4d5 [400]

3 years ago

6

Piper works at a camera store. He paid an hourly rate plus 16% commission on everything he sells. One week, he paid $365 for wor

king 20 hours and selling $1,500 worth of camera equipment. What is his hourly rate? *

1 answer:

Vlada [557]3 years ago

3 0

+5
1500 * 0.16 = 240
365 - 240 = 125
125 / 20 = 6.25 per hour

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Seth has a bank account which pays 1.01% interest, compounded quarterly. Seth withdraws $4,567 from the account every quarter fo

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Answer:

C. $538,021.66

Step-by-step explanation:

It is given that the money Seth withdraws was compounded every quarter for 35 years. So, we get,

Amount withdrawn every quarter, P = $4567

Rate of interest, r = $\frac{0.0101}{4}$ = 0.002525

Time period, n = 35 × 4 = 140

Now, as we know the formula for annuity as,

$P=\frac{r \times PV}{1-(1+r)^{-n}}$

where P = installments, PV = present value, r = rate of interest and n = time period.

This gives, $PV=\frac{P \times [1-(1+r)^{-n}]}{r}$

i.e. $PV=\frac{4567 \times [1-(1+0.002525)^{-140}]}{0.002525}$

i.e. $PV=\frac{4567 \times [1-(1.002525)^{-140}]}{0.002525}$

i.e. $PV=\frac{4567 \times [1-0.7021]}{0.002525}$

i.e. $PV=\frac{4567 \times 0.2975}{0.002525}$

i.e. $PV=\frac{1358.68}{0.002525}$

i.e. $PV=538,091.08$

So, the closest answer to initial value of the account is $538,021.66

Hence, option C is correct.

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Is the dilation from A to B a reduction or enlargement?

WITCHER [35]

Answer:

This is actually a reduction, and the scale factor of this dilation is 0.5.

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2 years ago

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Which of the following statements must be true about the polynomial function f(x)? If 1+ the sqrt of 13 is a root of f(x), then

son4ous [18]

The correct answer is: if $1 + 13i$ is a root of $f(x)$ , then $1 - 13i$ is also a root of $f(x)$ .

In fact, every polynomial has real and/or complex solutions. If all solutions are real, we're good. But if not all of them are real, then the complex ones come in couple of conjugate solutions. Since $1 + 13i$ and $1 - 13i$ are conjugate complex numbers, if one of them is a solution, the other must be as well.

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3 years ago

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<img src="https://tex.z-dn.net/?f=%28%20%5Csin%5E%7B2%7D%20%28%20%5Cfrac%7B%5Cpi%7D%7B%204%20%7D%20%20-%20%20%5Calpha%20%29%20%2

guapka [62]

Step-by-step explanation:

$\sin^2 (\frac{\pi}{4} - \alpha) = \frac{1}{2}(1 - \sin 2\alpha)$

Use the identity

$\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$

on the left side.

$\dfrac{1 - \cos [2(\frac{\pi}{4} - \alpha)]}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

$\dfrac{1 - \cos (\frac{\pi}{2} - 2\alpha)}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

Now use the identity

$\sin \theta = \cos(\frac{\pi}{2} - \theta)$

on the left side.

$\dfrac{1 - \sin 2\alpha}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

$\frac{1}{2}(1 - \sin 2\alpha) = \frac{1}{2}(1 - \sin 2\alpha)$

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2 years ago

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anyanavicka [17]

X=1 & Y=12
X=2 & Y=20
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3 years ago