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ExtremeBDS [4]
3 years ago
13

Zack is buying a new car with a base price of $16,750.she must also pay options, fees, and taxes shown on the bottom. the car de

alership will give her 48 months to pay off the entire amount. Yolanda can only afford to pay $395 each month. will she be able to buy the car? explain.
base price-$16,750
options-$500
fees-$370
taxes-$1,425
Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
5 0
It will take her 48 months and 2 weeks to pay
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Please help me with these two math questions!!
Westkost [7]

Answer:

the answer would be 1/6

Step-by-step explanation:

The reason for this is because 1/12 if you simplify 1/12 you would get 1/6. the number 2 is the same answer

5 0
4 years ago
Please help and thank you
Lorico [155]

Answer:

B. g(x) = (x + 4)²

Step-by-step explanation:

f(x) = x²

And since graph g(x) is f(x) moved 4 points along x-axis and to the right,

g(x) will be equal to (x + 4)²

g(x) = (x + 4)²

4 0
3 years ago
Find the simple interest on 500 at 6% for 1 year
ololo11 [35]

Answer:

Interest = $30

Step-by-step explanation:

Formula:

I = prt or I = p · r · t

So:

I = p r t

I = ($500)(.06)(1) <--- Just multiply all of these numbers and you get the final answer of:

Interest = $30.00 or just $30

8 0
3 years ago
Which number is divisible by 3?<br> A) 1,794<br> B) 1,912<br> C) 1,270<br> D) 473
Karolina [17]

Answer:

Step-by-step explanation:

should be A.

5 0
3 years ago
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
3 years ago
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