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3 years ago

How would I simplify this expression 7m^4/21m^5

1 answer:

4 0

the awnser to your problem would be 3m to the 9th power because 21m divided by 7m = 3m and then add the powers together so its 3m to the 9th power

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Leah has The flowers shown she wants to put them in 4 vases with the same number in each case how many flowers does she need to

dsp73

Answer:

The solution is obtained by dividing the number of flowers by the number of vases.

Step-by-step explanation:

The story problem is very straightforward. Normally, you need to read the problem and understand it.

Let's look at the question again.

Although we do not have all the quantities, we can still show how to solve the problem.

Let x be the total number of flowers.

There are 4 vases.

Therefore, the number of flowers in each vase will be:

x/4

Flow the same rule for similar problems.

5 0

3 years ago

N=3x45 write what n represents in the related division problem

Hitman42 [59]

One hundred thirty five 13%

6 0

3 years ago

Read 2 more answers

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago

HELP ME PLZ WILL MARK BR ASAP

Murljashka [212]

Answer:

the answer is A. 47

Step-by-step explanation:

tanθ = 15/14

the answer for 15/14 comes out to be about 1.0714 and so you put that value into your calculator using the sin^-1 button. it should look something like this sin^-1(1.0714) and press enter and you'll get 46.974 and round will result in 47

3 0

3 years ago

Question Progress

kozerog [31]

5 - 2 times 38 is -71.
Have a nice day

7 0

3 years ago

Read 2 more answers