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3 years ago

Which of the following are equivalent to the function y = 4cos x - 2? Check all that apply.

2 answers:

8 0

The answer is
<span>D. y = 4cos(-x) - 2
proof
cosine is an even function, it means cos(-x) = cosx
so </span>y = 4cos(-x) - 2 = y = 4cos(x) - 2

valina [46]3 years ago

6 0

Answer:

According to Apex, the answer is B and D.

Step-by-step explanation:

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Gre4nikov [31]

Answer:

$6.75

Step-by-step explanation:

15% of 45 = $6.75

5 0

3 years ago

One side of a rectangle is 20 cm larger than the other side. If you make the smaller side two times larger and the larger side t

grandymaker [24]

Let L be the length
Let w be the width
Let p be the perimeter
L+w+L+w=p
L=w+20
3L+2w+3L+2w=240
Sub the first equation in for L in the second equation and solve for w
3(w+20)+2w+3(w+20)+2w=240
3w+60+2w+3w+60+2w=240
10w+120=240
10w=240-120
10w=120
W=120/10
W=12
Sub w into the first equation and solve for L
L=w+20
L=12+20
L=32
Hope this helps!

3 0

3 years ago

You put $2,000 into a tax-deferred retirement account this year. your marginal tax rate is 15%. how much did you save on this ye

viva [34]

I want to say 150 which is answer A

6 0

3 years ago

Read 2 more answers

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago

Hugh sold 4 cars during a 5 hour shift, and later sold 12 cars sold during a 15 hour shift.

valentina_108 [34]

Answer:

Yes they are proportional because he is selling cars at the same rate

Step-by-step explanation:

7 0

3 years ago