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3 years ago

What is the simplest form of

Mathematics

1 answer:

ASHA 777 [7]3 years ago

8 0

Step-by-step explanation:

step 1. the cube root is also to the 1/3 power

step 2. (27a^3b^7)^(1/3) = (3*3*3*a*a*a*b*b*b*b*b*b*b)^(1/3)

step 3. a third root: 3 factors in the root comes "out" of the root as 1

step 4. 3ab*b(b)^(1/3) = 3ab^2(b)^(1/3).

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I need help for number 8 is it A B C or D

gizmo_the_mogwai [7]

The answer is a hope it helps

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3 years ago

In September, a website had 12,629 visitors. In October, the website had 24,290 more visitors than it did in September. The webs

Dvinal [7]

Answer:

the answer is 71,209 visitor

Step-by-step explanation:

12,629 + 24,290 + 24,290 = 71,209

7 0

3 years ago

PLEASE HELP me ANSWER these!!! My grade depends on it!!

iren [92.7K]

M (Mean) = 20,500;
SD (Standard Deviation) = 55
1 ) 20,555 = 20,500 + 55 = M + 1 SD
Answer: 50% + 34.1% = 84.1%
2 ) 20.610 = 20,000 + 110 = M + 2 SD
Answer: 10% - ( 50% + 34.1% + 13.6% ) = 100% - 97.7% = 2.3%
3 ) 20,445 = 20,500 - 55 = M - 1 SD
Answer: 50% - 34.1% = 15.9%

8 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago

What is the number three thousand eighty expressed in scientific notation?

valentinak56 [21]

Answer:

3.08 x $10^{3}$

Step-by-step explanation:

3080 = 3.08 x 1000 = 3.08 x $10^{3}$

6 0

3 years ago