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JulsSmile [24]
2 years ago
12

Find the product. (y3)^2 · y^7

Mathematics
2 answers:
Burka [1]2 years ago
7 0

Answer:

y^13

i got it right in my exam

Sedbober [7]2 years ago
6 0
Hey there! :D

When powers are outside the parenthesis, you multiply them. When they are not, you add them.

(y^3)^2 * y^7

3*2= 6

y^6*y^7

Add the powers. 

y^13= the simplest form

I hope this helps!
~kaikers
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3 years ago
For f(x) = 5x + 1
drek231 [11]

Answer:

A. f(7) = 5(7) + 1\\\\f(7) = 36

B. f^{-1}(x) = \frac{x - 1}{5}

C. f^{-1}(7) =\frac{6}{5}

D. f(\frac{6}{5}) = 7

Step-by-step explanation:

A. To solve the first part of the problem we must replace x = 7 in the function f(x) = 5x + 1

So:

f(7) = 5(7) + 1\\\\f(7) = 36

B. In part B we must find the inverse function of f(x) = 5x + 1

To find the inverse function do y = f(x)

y = 5x +1

Now clear the variable x.

\frac{y - 1}{5} = x

Replace x with y.

y = \frac{x - 1}{5}

Finally

f^{-1}(x) = \frac{x - 1}{5}

C. Now we take the inverse function found above and replace x = 7

f^{-1}(7) = \frac{7 - 1}{5}\\\\f(7) = \frac{6}{5}

D. Now we substitute x = f(f^{-1}(7)) in the original function.

x = f( f^{-1}(7))\\\\f^{-1}(7) = \frac{6}{5}\\\\ x= f(\frac{6}{5} )\\\\f(\frac{6}{5}) = 5(\frac{6}{5}) + 1\\\\f(\frac{6}{5}) = 7

8 0
3 years ago
Read 2 more answers
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Viktor [21]

Answer:

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Step-by-step explanation:

<u>Theorem: The diagonals of a kite are perpendicular.</u>

Let O be the point of intersection of the diagonals,

Applying Pythagoras Theorem, in right triangle WOX

|WX|^2=|WO|^2+|OX|^2\\|WX|^2=3^2+3^2=18\\|WX|=\sqrt{18} =3\sqrt{2}\:Units \:or\: 4.2 \:Units

Applying Pythagoras Theorem, in right triangle WOZ

|WZ|^2=|WO|^2+|OZ|^2\\|WX|^2=3^2+5^2=34\\|WZ|=\sqrt{34} \:Units \:or\: 5.8 \:Units

4 0
2 years ago
Sarah jogged 4.8 miles each day for 22 days last month. How many miles did Sarah jog in total?
solong [7]

Answer:

105.6

hope this helps

have a good day :)

Step-by-step explanation:

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elena55 [62]

Answer:

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4 0
2 years ago
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