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Elodia [21]
3 years ago
7

How many fourths are in 2/3? NEED ANSWER ASAP!!!

Mathematics
2 answers:
natima [27]3 years ago
5 0
I got 2.6 I hope that's right
olga nikolaevna [1]3 years ago
4 0
There are 2.67 fourths in 2/3 cups
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Solve the equation for x.<br> 12x + 20 = 5x - 10 - 3x
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Answer:

x = -3

Step-by-step explanation:

seperate this equation into different sides where like terms are on each side.

12x + 20 = 5x - 10 - 3x

12x - 5x + 3x = -10 - 20

simplify

10x = - 30

x =  - 3

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What is the initial value for the function f(x)=3(5/4)x
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For
f(x)=3\left(\dfrac{5}{4}\right)^{x}
you seem to want to know the value of f(0). Put 0 where x is, then evaluate.
f(0)=3\left(\dfrac{5}{4}\right)^{0}=3\cdot 1=3

The initial value of the function is 3.

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You will observe that when x=0, the exponential term evaluates to 1. This is true regardless of its base. Thus f(0) is always the coefficient of the exponential term—in this case, 3.
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A rectangular box with a volume of 684 ftcubed is to be constructed with a square base and top. The cost per square foot for the
SCORPION-xisa [38]

Answer:

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost=2046.16 cents      

Step-by-step explanation:

Given that a rectangular box with a volume of 684 ft³.

The base and the top of the rectangular box is square in shape

Let the length and width of the rectangular box be x.

[since the base is square in shape,  length=width]

and the height of the rectangular box be h.

The volume of rectangular box is = Length ×width × height

                                                      =(x²h) ft³

x^2h=684\Rightarrow h=\frac{684}{x^2}  (1)

The area of the base and top of rectangular box is = x² ft²

The surface area of the sides= 2(length+width) height

                                               =2(x+x)h

                                               =4xh ft²

The total cost to construct the rectangular box is

=[(x²×20)+(x²×15)+(4xh×1.5)] cents  

=(20x²+15x²+6xh) cents

=(25x²+6xh) cents

Total cost= C(x).

C(x) is in cents.

∴C(x)=25x²+6xh

Putting h=\frac{684}{x^2}

C(x)=25x^2+6x\times\frac{684}{x^2} \Rightarrow C(x)=25x^2+\frac{4104}{x}

Differentiating with respect to x

C'(x)=50x-\frac{4104}{x^2}

To find minimum cost, we set C'(x)=0

\therefore50x-\frac{4104}{x^2}=0\\\Rightarrow50x=\frac{4104}{x^2}\\\Rightarrow x^3=\frac{4104}{50}\Rightarrow x\approx 4.345 ft.

Putting the value x in equation (1) we get

h=\frac{684}{(4.345)^2}

 ≈36.23 ft.

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost C(x)=[25(4.345)²+10(4.345)(36.23)] cents

                               =2046.16 cents      

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