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KengaRu [80]
2 years ago
15

Which equation results from isolating a radical term and squaring both sides of the equation for the equation

Mathematics
1 answer:
astraxan [27]2 years ago
8 0

Answer:

c - 2 = 25 + 10 \sqrt c + c

Step-by-step explanation:

Given

\sqrt{c-2} - \sqrt c = 5

Required

Isolate radical, then square both sides

We have:

\sqrt{c-2} - \sqrt c = 5

Isolate radical

\sqrt{c - 2} = 5 + \sqrt c

Square both sides

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

c - 2 = 25 + 2 * 5 * \sqrt c + c

c - 2 = 25 + 10 \sqrt c + c

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2 years ago
Ma.clark spends $89.85 on science kits that cost $5 each plus $4.85 tax on the total bill. How many kits did she buy?
k0ka [10]
The answer to this question is 17
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3 years ago
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1) g(n)= 3n - 4; Find g(-3)​
nalin [4]

Answer:

g(-3) = -13

Step-by-step explanation:

Plug in -3 for n in the equation:

g(n) = 3n - 4

g(-3) = 3(-3) - 4

Remember to follow PEMDAS. First, multiply, then subtract:

g(-3) = 3 * -3 = -9

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-13 is your answer for g(-3).

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3 years ago
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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

                                           =x^2h

Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

=20x^2 cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

=4xh\times 1.5 cents

=6xh cents

Total cost = (20x^2+10x^2+6xh)

                =30x^2+6xh

Let

C=30x^2+6xh

Putting the value of h

C=30x^2+6x\times \frac{272}{x^2}

\Rightarrow C=30x^2+\frac{1632}{x}

Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

                                          =30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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3 years ago
Find the general solution: y''-2y'-3y=-3te^(-t)
allsm [11]
The answer i got was -3te with -t as an exponent 
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3 years ago
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