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Lina20 [59]
3 years ago
15

Amanda is going to rent a truck for one day. There are two companies she can choose from, and they have the following prices.

Mathematics
1 answer:
nydimaria [60]3 years ago
5 0

Answer:

The mileages for which Company A charges less than Company B is 0 miles ≤ m ≤ 85 miles

Step-by-step explanation:

The given parameters for the two companies are;

For Company A

Company A initial fee = $55

The additional fee per mile = 70 cents

For Company B

Company B initial fee = $97.50

The additional fee per mile = 20 cents

Let C_A represent the total charge for company A and let C_B represent the total charge for company B, we have;

C_A = 55 + 0.70 × m

C_B = 97.50 + 0.20 × m

Where "m" represents the number of miles driven

Therefore, the cost of Company A and Company B are the same or equal when have the following relation;

C_A = C_B

∴ 55 + 0.70 × m = 97.50 + 0.20 × m

0.70 × m - 0.20 × m = 97.50 - 55

0.5 × m = 42.5

m = 85

The number of miles driven for which the two companies will charge the same amount, m = 85 miles

Therefore, given that the initial fee for Company A is less than the initial fee for Company B. we have;

Therefore for 0 miles  ≤ m < 85 miles, Company A charges less than Company B.

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Glven: F(x) = 2x - 1; G(X) = 3x + 2; H(x) = x2<br> Find F/G/H(2)<br> 121<br> 27<br> 71
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Step-by-step explanation:

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Then (F/G)(2) = (3/8), and (F/G/H) = -------- = 3/32

                                                              4

8 0
3 years ago
Stacey has a square piece of cloth. She cuts 3 inches off of the length of the square and 3 inches
Mamont248 [21]

Answer:

6 in

Step-by-step explanation:

Let x = the side length of the original square.

They removed 3 in from each side of the original square, so the side lengths of the remaining square are x - 3 in.

The area of the smaller square is (x - 3)².

The area of the original square is x²

I assume the area of the smaller square is ¼ that of the original square. Then

1. Solve for x

\begin{array}{rcl}\frac{1}{4}x^{2} & = & (x - 3)^{2}\\x^{2} & = & 4(x - 3)^{2}\\& = & 4(x^{2} - 6x + 9)\\x^{2}& = & 4x^{2} - 24x + 36\\3x^{2} - 24x + 36 & = & 0\\x^{2} - 8x + 12 & = & 0\\(x - 2)(x - 6) & = & 0\\x - 2 = 0& \qquad &x - 6 = 0\\x = 2& \qquad &x = 6\\\end{array}

2. Calculate the side length of the smaller square

(a) x = 2

Side length = x - 3 = 2 - 3 = -1 in.

IMPOSSIBLE. You can't have a negative side length.

(b) x = 6

Side length of smaller square = 6 - 3 = 3 in.

Side length of original square = x = 6 in

Check:

\begin{array}{rcl}\frac{1}{4}(6)^{2} & = & (6 - 3)^{2}\\\frac{1}{4}\times 36 & = & 3^{2}\\9 & = & 9\\\end{array}

OK.

3 0
3 years ago
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