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3 years ago

Write the following as a radical expression.

Mathematics

1 answer:

jeyben [28]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Using the rule of exponents/ radicals

$a^{\frac{m}{n} }$ = $\sqrt[n]{a^{m} }$

Given

$z^{\frac{3}{8} }$

= $\sqrt[8]{z^{3} }$

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The scale of a map is 1 in. :500 mile. City A is 650 miles from City B. How far is it's distance on the map?

crimeas [40]

Set up a proportion
1/500=x/650
Cross multiply
650=500x
Divide both sides by 500
1.3=x
So the distance on the map would be 1.3 inches

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tigry1 [53]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

In the diagram, AB = 12, DX = 2.5, and BX = 5. Find CD. Show all work.

Naily [24]

I found your complete question in another source.
See attached image.
For this case what you should see is that you have two similar triangles.
We then have the following relationship:
((AB) / (BX)) = ((CD) / (DX))
We cleared CD:
CD = ((AB) / (BX)) * (DX)
Substituting the values:
CD = ((12) / (5)) * (2.5)
CD = 6
Answer:
CD = 6

5 0

3 years ago

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

3 years ago

HELP I NEED HELP ASAP

V125BC [204]

Answer:

try C or A

Step-by-step explanation:

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3 years ago