Please help me!

Mathematics

1 answer:

Temka [501]3 years ago

7 0

7. Answer is D! 1 to the 3rd power, 1^3, or 1 cubed as you'll hear it, is 1 because when something is cubed that means it's times itself 3 times. (EX: 4^3 = 4x4x4) 2 cubed is 8. 3 cubed is 27. 4 cubed is 64 and 5 cubed is 125!

8. Answer is C! A is gaining speed and B is the capped speed, meaning it can't get any higher, and C is the loss of speed.

You might be interested in

the length of a rectangle is 4/5 feet the width is 5/8 feet how much greater is the length of a rectangle than the width

Eva8 [605]

4/5(8)= 32/40
5/8(5)= 25/40
You have to get the same denominator to be able to subtract so multiply one fractions denominator by the other so it becomes the same, and vise versa.
32/40-25/40= 7/40 which cannot be simplified.
Answer= 7/40 greater than the width.

3 0

4 years ago

What is 9 times 2/8 equal ? Please help

liberstina [14]

It equals 2.25 because you multiple 2 and 9 and then divide the answer by 2.

6 0

3 years ago

Read 2 more answers

The table represents the start of the division of by the indicated divisor. What is the quotient?

arlik [135]

The answer is the quotient

5 0

3 years ago

Read 2 more answers

Answer this question correctly, please give the correct answer.

Advocard [28]

X = -2

set equations equal and solve
2x+7 = -x+1
collect like terms
3x = -6
divide
x=-2

7 0

3 years ago

Use logarithmic differentiation to find dy/dx: y=((x^3)(2x+3)^1/2) / (x-2)^2

sukhopar [10]

$y=\frac { { { x }^{ 3 }\left( 2x+3 \right) }^{ \frac { 1 }{ 2 } } }{ { \left( x-2 \right) }^{ 2 } }$

$\\ \\ { \left( x-2 \right) }^{ 2 }\cdot y={ x }^{ 3 }{ \left( 2x+3 \right) }^{ \frac { 1 }{ 2 } }$

$\\ \\ \ln { \left( { \left( x-2 \right) }^{ 2 }\cdot y \right) } =\ln { \left( { x }^{ 3 }{ \left( 2x+3 \right) }^{ \frac { 1 }{ 2 } } \right) }$

$\\ \\ \ln { \left( { \left( x-2 \right) }^{ 2 } \right) } +\ln { y } =\ln { \left( { x }^{ 3 } \right) } +\ln { \left( { \left( 2x+3 \right) }^{ \frac { 1 }{ 2 } } \right) }$

$\\ \\ 2\ln { \left( x-2 \right) } +\ln { y } =3\ln { x } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right) }$

$\\ \\ \ln { y } =3\ln { x } -2\ln { \left( x-2 \right) } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right) }$

$\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }$

$\\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 } \right)$

$\\ \\ \frac { dy }{ dx } =\frac { { x }^{ 3 }\sqrt { 2x+3 } }{ { \left( x-2 \right) }^{ 2 } } \cdot \left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 } \right)$

4 0

3 years ago

Read 2 more answers