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ella [17]
3 years ago
5

Lester’s car can go 15.4 miles on 1 gallon of gas. How far can he go on 0.7 gallon?

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

He can go 10.78 miles

Step-by-step explanation:

This problem can be solve by creating a proportion, knowing that 15.4 miles is to one gallon the same as "x" miles (our unknown) is to 0.7 gallons. This can be set in math terms as the equation that relates these two ratios of miles per gallon:

\frac{15.4\,miles}{1\,gallon} =\frac{x }{0.7\,gallon}

and to solve for the unknown number of mile, we multiply both sides of the equation by "0.7 gallon" in order to isolate our unknown:

\frac{15.4\,miles}{1\,gallon} =\frac{x}{0.7\,gallon}\\\frac{15.4\,miles\,*\,0.7\,gallon}{1\,gallon} =x \\x=10.78\,miles

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A bird leaves its nest and travels 12 miles per hour downwind for x hours. On the return trip, the bird travels 2 miles per hour
aalyn [17]
12x = 10x + 3

Subtract 10x from both sides

2x = 3
Divide by two on each side

x = 3/2 or 1 1/2 hours



Distance of the trip:

12 * 2 * 3/2 = 36 miles

How long does the trip take:

3/2 + 3/2 + 3/10 = 3 3/10 hours
6 0
3 years ago
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Answer:

5^{12}

Step-by-step explanation:

Lets say there's a number 'x' and there's an expression (x^a)^b where 'a' & 'b' are also numbers , <u>the simplified form of the expression is </u>x^{a \times b}<u>.</u>

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2 years ago
1. Carter will install fencing all around the flat area of his
MariettaO [177]

Answer:

Amount of fencing required = 75 yards

Step-by-step explanation:

Distance between the two points (x_1,y_1) and (x_2,y_2) is given by the formula,

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance between A(3, 6) and B(3, -2) = \sqrt{(3-3)^2+(6+2)^2}

                                                               = 8 yards

Distance between B(3, -2) and C(-7, 4) = \sqrt{(3+7)^2+(-2-4)^2}

                                                                = \sqrt{136}

                                                                = 11.66 ≈ 12 yards

Distance between C(-7, 4) and D(-7, -2) = \sqrt{(-7+7)^2+(4+2)^2}

                                                                 = 6 yards

Distance between D(-7, -2) and E(-3, -2) = \sqrt{(-7+3)^2+(-2+2)^2}

                                                                   = 4 yards

Distance between E(-3, -2) and F(-3, -8) = \sqrt{(-3+3)^2+(-2+8)^2}

                                                                = 6 yards

Distance between F(-3, -8) and G(3, -8) = \sqrt{(-3-3)^2+(-8+8)^2}

                                                                 = 6 yards

Distance between G(3, -8) and H(10, -12) = \sqrt{(3-10)^2+(-8+12)^2}

                                                                   = \sqrt{49+16}

                                                                   = \sqrt{65}

                                                                   = 8.06 ≈ 8 yards

Distance between H(10, -12) and J(10, 6) = \sqrt{(10-10)^2+(-12-6)^2}

                                                                   = 18 yards

Distance between A(3, 6) and J(10, 6) = \sqrt{(10-3)^2+(6-6)^2}

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Since length of fence required = perimeter of the flat area

Perimeter of the given area = 8 + 12 + 6 + 4 + 6 + 6 + 8 + 18 + 7

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Therefore, amount of fencing required = 75 yards

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