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Ksivusya [100]
3 years ago
10

What is x times the quotient of y times z in number form?

Mathematics
1 answer:
morpeh [17]3 years ago
8 0
Are you missing anything else from the question?
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A sheet of paper has a perimeter of 39 in. if the width of the paper is 8.5 in . what is the length of the sheet of paper
Valentin [98]
If the perimeter is 39 what you have to do is 39 - 8.5 - 8.5 that will give you 22 and divide 22/2 and that gives you 11. So 11 is the length
5 0
3 years ago
Read 2 more answers
I need help or I’m going to fail math
gavmur [86]

Answer:

1. A

2. B

3. D

Step-by-step explanation:

Use PEMDAS to solve all

1.

8^2 - 6*2

8*8 - 6*2

64 - 12 = 52

2.

4^2 * 3 + 4 * 2

4 * 4 * 3 + 4 * 2

16 * 3 + 8

48 + 8 = 56

3.

9^2 - 2(5+3)

9 * 9 - 2(8)

81 - 16 =65

4 0
2 years ago
Figure B is a scaled copy of Figure A.Select all of the statements that must be true:
KatRina [158]

Answer:

B) Figure B has the same number of edges as Figure A

D) Figure B has the same number of angles as Figure A

E) Figure B has angles with the same measures as Figure A

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

In this problem

If Figure B is a scaled copy of Figure A

then

Figure A and Figure B are similar

therefore

<u><em>The statements that must be true are</em></u>

B) Figure B has the same number of edges as Figure A

D) Figure B has the same number of angles as Figure A

E) Figure B has angles with the same measures as Figure A

6 0
3 years ago
Can someone please solve this please.
Lelu [443]

Answer:

A≈576.31

Step-by-step explanation:

surface area = bh + L(s1+s2+s3)

surface area = (13*11) + 13 * (10+11+14.87)

8 0
2 years ago
Find all the solutions for the equation:
Contact [7]

2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0

Divide both sides by x^2\,\mathrm dx to get

2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}

Substitute v(x)=\dfrac{y(x)}x, so that \dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}. Then

x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}

The remaining ODE is separable. Separating the variables gives

\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x

Integrate both sides. On the left, split up the integrand into partial fractions.

\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}

\implies v^2+2v+1=a(v^2+1)+(bv+c)v

\implies v^2+2v+1=(a+b)v^2+cv+a

\implies a=1,b=0,c=2

Then

\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v

On the right, we have

\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C

Solving for v(x) explicitly is unlikely to succeed, so we leave the solution in implicit form,

\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C

and finally solve in terms of y(x) by replacing v(x)=\dfrac{y(x)}x:

\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}

7 0
3 years ago
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