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Bogdan [553]
2 years ago
10

Can someone help me with a quick Algebra question, Thanks! ;)

Mathematics
1 answer:
Viktor [21]2 years ago
3 0
The answer is equal to (B). The solutions are x = -7.7 and x = -1.3.
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The square root of -15 is 3.87298335
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Write answer as x=<br><br> x-21=99
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X=120
This is the answer because 120-21=99
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Write an expression that evaluates to true if the int associated with number_of_prizes is divisible (with no remainder) by the i
hodyreva [135]

We need to find the expression for " number_of_prizes is divisible number_of_participants". Also there should not remain any remainder left. On in order words, we can say the reaminder we get after division is 0.

Let us assume number of Prizes are = p and

Number of participants = n.

If we divide number of Prizes by number of participants and there will be not remainder then there would be some quotient remaining and that quotent would be a whole number.

Let us assume that quotent is taken by q.

So, we can setup an expression now.

Let us rephrase the statement .

" Number of Prizes ÷ Number of participants  = quotient".

p ÷ n = q.

In fraction form we can write

p/n =q   ; n ≠ 0.


4 0
3 years ago
Light travels about 186,000 mi/s, the function d(t)=186,000 gives the distance in miles that light travels in T seconds , how fa
I am Lyosha [343]
Light travels 5,580,000 mi/s in thirty seconds
4 0
3 years ago
A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo
katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the boxV=x^2z

Volume of the box=12 ft^3\\

Therefore, x^2z=12\\z=\frac{12}{x^2}

The material for the base costs \$0.17/ft^2, the material for the sides costs \$0.10/ft^2, and the material for the top costs \$0.13/ft^2.

Area of the base =x^2

Cost of the Base =\$0.17x^2

Area of the sides =4xz

Cost of the sides==\$0.10(4xz)

Area of the Top =x^2

Cost of the Base =\$0.13x^2

Total Cost, C(x,z) =0.17x^2+0.13x^2+0.10(4xz)

Substituting z=\frac{12}{x^2}

C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}

To minimize C(x), we solve for the derivative and obtain its critical point

C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2

Recall: z=\frac{12}{x^2}=\frac{12}{2^2}=3\\

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0
3 years ago
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