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Delicious77 [7]
3 years ago
9

Prove that x=2 if 4^x=8x...where ^ stands for "to the power of".

Mathematics
1 answer:
Romashka [77]3 years ago
4 0
In order to prove this, you simply plug in the number 2 everywhere you see X:

4^(2) = 8(2). 

Simplify

16 = 16 √

Since this checks out, x is proven to be 2.
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Answer:

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Work backward to find the value of the variable in the equation below d-4=23
AnnyKZ [126]

Answer:

27

Step-by-step explanation:

The equation is a onestep equation, so in order to find the variable, follow these steps:

d - 4 = 23

In order to isolate the variable or get the variable by itself, you need to get rid of the constants around it or the numbers without a variable. In order to do that for this equation, you'll need to do the inverse of subtracting 4, which would be adding four to both sides.

Adding 4 to positive 23 will give you the answer of 27;

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4 years ago
Find the value of x and y.
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Answer:

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3 years ago
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Find the product. (-2x^2 )^3 ·3 x
jeka94

Assignment: \bold{Solve \ \left(-2x^2\right)^3\cdot \:3x}

<><><><><>

Answer: \boxed{\bold{-24x^7}}

<><><><><>

Explanation: \downarrow\downarrow\downarrow

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[ Step One ] Rewrite \bold{\left(-2x^2\right)^3}

\bold{-2^3\left(x^2\right)^3}

[ Step Two ] Rewrite Equation

\bold{-2^3\cdot \:3x^6x}

[ Step Three ] Apply Exponent Rule

Note: \bold{Exponent \ Rule \ \:a^b\cdot \:a^c=a^{b+c}: \ x^6x=\:x^{6+1}=\:x^7}

\bold{-2^3\cdot \:3x^7}

[ Step Four ] Refine

\bold{-24x^7}

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3 years ago
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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
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