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PIT_PIT [208]
3 years ago
13

9 negative exponent 8

Mathematics
1 answer:
Yuliya22 [10]3 years ago
6 0

Answer:

See below:

Step-by-step explanation:

So we want to know what 9 to the -8th exponent will be if I understand your question right.

To solve for 9^{-8} we first have to solve for 9^{8} which would be 43046721.

Now, if we have a negative number, the answer would be 1 divided by that number, so therefore, we get 9^{-8} =\frac{1}{43046721}

Cheers!

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506 divided by 22 is 23. 
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(X+2) (x+8) <br> expand and simplify
Nutka1998 [239]

Answer:

X+2=2x

x+8=8x

(2x).(8x)=16x

8 0
3 years ago
Find the area of this circle. Use 3 for a.A = 7r2=11 in[?] in?
liberstina [14]

SOLUTION

We want to find the area of the circle in the picture given.

We have been given the formula as

\begin{gathered} A=\pi r^2 \\ We\text{ are told to take }\pi\text{ as 3} \\  \end{gathered}

From the circle, the radius r = 11 in.

The area of the circle becomes

\begin{gathered} A=\pi r^2 \\ A=\pi\times r^2 \\ A=3\times11^2 \\ A=3\times11\times11 \\ A=363in^2 \end{gathered}

Hence the answer is 363 square-inches

7 0
1 year ago
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much
Y_Kistochka [10]

Answer:

x = ∛ 2*V/5  

y = ∛ 2*V/5

h  = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is  x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h      where   h =  V / x*y

As the base is a rectangular one there are 2 sides x*h .  and 2 sides  y*h

According to this:

Ct (cost of aquarium )  = cost of the base  + cost of the sides

cₐ  ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h   +   c₁*2*y*h

C(t)  =  5*c₁*x*y +2* c₁*x* V/x*y  +  2* c₁*y* V/x*y    or

C(t)  =  5*c₁*x*y  + 2*c₁*V/y   *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x)  =  5*c₁*y - 2*c₁*V/x²

C´(y)  =  5*c₁*x - 2*c₁*V/y²

C´(x)  = C´(y)        ⇒  5*c₁*y - 2*c₁*V/x²  =   5*c₁*x - 2*c₁*V/y²

or    5*y - 2*V/x²  =   5*x - 2*V/y²

(5*y*x² - 2*V)/x²  = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y²  = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y²  =  5*y²x³  - 2*V*x²

5*y³*x² - 5*y²x³  =  2*V * ( y² - x²)

by symmetry  x =  y

Then using x = y  and plugging that value on the derivatives

C´(x) =  5*c₁*y - 2*c₁*V/x²

C´(x) =  5*c₁*x - 2*c₁*V/x²

C´(x) = 0          ⇒     5*c₁*x - 2*c₁*V/x²  = 0

5*x  - 2*V/x² = 0      ⇒  5*x³ - 2*V = 0   ⇒   5*x³  = 2*V  ⇒ x³ = 2*V/5

x = ∛ 2*V/5       and   y = ∛ 2*V/5    and   h  =  V/ x*y    h  = V/ ∛ 4*V²/25

7 0
3 years ago
Anyone mind helping me out?
Airida [17]
<h2>6 is the answer.</h2>

just simple

3 0
3 years ago
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