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3 years ago

Whats the answer to x^3 y^0 z^-2

Mathematics

1 answer:

Elina [12.6K]3 years ago

5 0

Answer:

$\frac{x^3}{z^2}$

Step-by-step explanation:

There are no given values, therefore we can't determine a definite answer. However, we are able to simplify the expression.

As we all know that any variable or number to the power of 0 is equal to 1. This means that $y^0 = 1$ .

We also know that a negative power can bring the variable down and make it a fraction. This means that $z^-^2=\frac{1}{z^2}$

Now that we got these two sorted we are able to write it in a simplified form.

Answer: $\frac{x^3}{z^2}$

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FOR 12 POINTS I NEED HELP ASAP! Consider the system of linear equations. 5 x + 10 y = 15. 10 x + 3 y = 13 To use the linear comb

Romashka [77]

Answer:

Do you mean which equation should the 5x+10=15 be multiplied to?

it would be -2

Step-by-step explanation:

This is because 5 times -2 is -10

and so when you add the (now) x value of the first equation to the x value of the second equation, it gets 0

which makes it eliminated!

-10x+10x=0

8 0

3 years ago

Show that (x-5) is a factor of x^3-3x^2-13x+15

mario62 [17]

Answer:

sorry not in that grade

Step-by-step explanation:

6 0

3 years ago

Solve the equation for b.

Mkey [24]

Answer:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

b = M /4 + 9 x

7 0

2 years ago

Read 2 more answers

Which expressions are equivalent to -2(5x-3/4)?

kogti [31]

Answer:

-10x+3/2

Hope this helps

6 0

2 years ago

After being treated with chemotherapy, the radius of the tumor decreased by 23%. What is the corresponding percentage decrease i

Crazy boy [7]

Answer:

The volume of the tumor experimented a decrease of 54.34 percent.

Step-by-step explanation:

Let suppose that tumor has an spherical geometry, whose volume ( $V$ ) is calculated by:

$V = \frac{4\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the tumor.

The percentage decrease in the volume of the tumor ( $\%V$ ) is expressed by:

$\%V = \frac{\Delta V}{V_{o}} \times 100\,\%$

Where:

$\Delta V$ - Absolute decrease in the volume of the tumor.

$V_{o}$ - Initial volume of the tumor.

The absolute decrease in the volume of the tumor is:

$\Delta V = V_{o}-V_{f}$

$\Delta V = \frac{4\pi}{3}\cdot (R_{f}^{3}-R_{o}^{3})$

The percentage decrease is finally simplified:

$\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%$

Given that $R_{o} = R$ and $R_{f} = 0.77\cdot R$ , the percentage decrease in the volume of tumor is:

$\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%$

$\%V = 54.34\,\%$

The volume of the tumor experimented a decrease of 54.34 percent.

5 0

2 years ago