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Varvara68 [4.7K]
4 years ago
8

(8x^9y^3)^-2 (6x^7y^7) simplify. write the answer using positive exponents only

Mathematics
1 answer:
SOVA2 [1]4 years ago
4 0
<span>(8x^9y^3)^-2 (6x^7y^7) simplify. write the answer using positive exponents only

</span>(8x^9y^3)^{-2} (6x^7y^7)=  \\  \\  \\ (8)^{-2} (x^9) ^{-2}( y^3)^{-2} (6)(x^7)(y^7)=  \\  \\  \\ (8)^{-2} (x) ^{9-2}( y)^{3-2} (6)(x^7)(y^7)=  \\  \\  \\ (8)^{-2} (x) ^{9-2+7}( y)^{3-2+7} (6)=  \\  \\  \\ 6*(8)^{-2} (x) ^{14}( y)^{8} = \\  \\  \\   \dfrac{6}{64} x ^{14} y^{8} = \boxed{\dfrac{3}{32}\  x ^{14} y^{8} }<span>
</span>
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Simplify the expression by<br> distributing, then combining like<br> terms!<br> 4 (3x + 2) - 6
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Answer:

\boxed {12x + 2}

Step-by-step explanation:

Solve the following expression:

4(3x + 2) - 6

-Use <u>Distributive Property</u>:

4(3x + 2) - 6

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-Combine like terms:

12x + 8 - 6

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4 0
3 years ago
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skad [1K]

Answer:

Therefore the value of perimeter of the rectangle is 14 in.

Step-by-step explanation:

The opposite sides of a rectangle are congruent.

The perimeter of a rectangle is 2(length + width)

The area of a rectangle is (length×width).

Given the vertices of a rectangle are P(2,2), Q(6,2), R(6,5) and S(2,5).

To find the the perimeter, first we need the sides of the rectangle.

The distance between two points(x₁,y₁) and (x₂,y₂) is

=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

The distance between P and Q is

=\sqrt{(6-2)^2+(2-2)^2}  in

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The distance between Q and R is

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=\sqrt{0^2+3^2} in

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The distance between R and S is

=\sqrt{(2-6)^2+(5-5)^2} in

=\sqrt{ (-4)^2} in

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The distance between S and P is

=\sqrt{(2-2)^2+(2-5)^2} in

=\sqrt{(-3)^2} in

=3 in

Therefore the length of the rectangle is = 4 in.

an the width of the rectangle is = 3 in

Therefore the perimeter of the rectangle is =2(4+3) in

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3 years ago
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Answer:

You taught for 4 hours

Step-by-step explanation:

Let this number of hours be x.  Then:

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Equate this to $52 and then solve the resulting equation for x:

20 + 8x = 52

Then 8x = 32, and x = 4.  You taught for 4 hours.

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Is this correct? 8th grade math
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