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V125BC [204]
2 years ago
6

Joe takes out a loan for $19,000. He will pay the loan back monthly. The interest rate is 7.6% per year and the length of the lo

an is 4.5 years. TEKS 8.12(A), 8.12(B) 1. Determine how many payments need to be made. Explain.
Mathematics
1 answer:
Y_Kistochka [10]2 years ago
6 0
I believe it should be 8,000
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I am M I am a number between 17 and 25 I am a multiple of 36 is not a factor of mine what am I
Leni [432]

Answer:

wat

Step-by-step explanation:

8 0
2 years ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
Jenny’s home appraised for 550,000. Her asking price was 560,000 and the properties sold for 575,000. What were the new owners p
Alisiya [41]

Answer:

575,000

Step-by-step explanation:

5 0
2 years ago
Amira is painting a rectangular banner 2 1/4 yards wide in the cafeteria. The banner will have a blue background. Amira has enou
elena55 [62]
Area of a rectangle:
A = w x h, where w is the width and h is the height.
w = 2.25 yd
A = 1.5 yd²
1.5 = 2.25 · h
h = 1.5 : 2.25 = 0.6667 = 2/3 yd
Answer:
The height of a banner is 2/3 yard.
4 0
3 years ago
Read 2 more answers
General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}
mixer [17]

Answer:

\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

Step-by-step explanation:

Given:

\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1

First, note that

\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x

So, the equation is

-\cos x+\sin x= -1

Multiply this equation by \frac{\sqrt{2}}{2}:

-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}

The general solution is

x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

4 0
3 years ago
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