MATHEMATICS HIGH SCHOOL

Lilianna discovered that the ratio of kilometers to miles is 1 to 0.6 if its 60 kilometers to Moncton then how many miles does it take to get to Moncton.

Answers

Answer 1
Answer:

Answer: 36 miles.

Step-by-step explanation: Ratio of 1 to 0.6 means 1:0.6 or (1)/(0.6).

So, if to Moncton is 60km:

(1)/(0.6)=(60)/(y)

y=60.0.6

y = 36

To Moncton, it is 36 miles.

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8. What is the solution to the system? y = 1/2x, 2x + 3y = 28A) (2, 3)
B) (8, 4)
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D) (2, 14)

Answers

y=(1)/(2)x \n2x+3y=28 \n \n\hbox{substitute } (1)/(2)x \hbox{ for y in the second equation:} \n2x+3 ((1)/(2)x)=28 \n2x+(3)/(2)x=28 \ \ \ \ \ |* 2 \n4x+3x=56 \n7x=56 \ \ \ \ \ \ \ \ \ \ \ \ |/ 7 \nx=8 \n \ny=(1)/(2)x=(1)/(2) * 8=4 \n \n\boxed{(x,y)=(8,4)} \Leftarrow \hbox{answer B}

Evaluate the expression 2x - (4y - 3) + 5xz, when x = -3, y = 2, and z = -1

Answers

Substitute each variable for the coefficient. Remember PEMDAS.

The revenue, in dollars, of a company that produces jeans can be modeled by 2x2 + 17x – 175. The cost, in dollars, of producing the jeans can be modeled by 2x2 – 3x – 125. The number of pairs of jeans that have been sold is represented by x. The profit is the difference between the revenue and the cost, 20x – 50.If 75 pairs of jeans are sold, what is the company’s profit?

Answers

To answer the problem above we apply the given. If x is represented by the number of jeans sold then we only need to substitute x. 20x - 50 = 20(75) - 50 = 1500 - 50 = 1450. Th profit of the company is 1450,

Answer:

B: $1450

Step-by-step explanation:

One positive number is 5 times another number. The difference between the two numbers is 200, find the numbers.

Answers

x = 5y, where x > 0 and y > 0;
x - y = 200 => 5y - y = 200 => 4y = 200 => y = 200/4 => y = 50 => x = 250.

Solve for x
64°
Х
45°

Answers

Answer:

x = 109

Step-by-step explanation:

64 + 45 = 109

180 - 109 = 71

180 - 71 = 109

Solve the linear equation for x

-4.8(6.3x-4.18) =-58.56

X=

Answers

Answer:

x=2.6

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.
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