Question

1. Jackson has 75¢ in dimes, d, and nickels, n, in his pocket. Which equation could be solved to find the possible combinations of dimes and nickels Jackson has?
A. 75 = d + n B. 75 = dn C. 75 = 10d ·5n D. 75 = 10d + 5n

The linear equation below has two variables. PLEASE HELP ME!
y = (-1/4)x – 1

Which shows the solutions to the equation when x = 4, x = 0, and x = -4?

A. (-2, 4), (1, 0), (0, -4) C. (2, -4), (-1, 0), (0, 4)
B. (4, -2), (0, -1), (-4, 0) D. (-4, 2), (0, 1), (4, 0)

A 50-foot roll of fencing will be used to enclose a rectangular garden. Which equation below could not be solved to find the possible lengths, l, and widths, w, of the garden?

A. 50 = 2lw C. 2(l + w) = 50
B. 50/2 = l + w D. 50 = 2l + 2w


4. Ms. Monti bought x adult tickets and y children’s tickets to an ice-skating show. She spent a total of $145. The equation below describes the relationship between x and y.

25x + 15y = 145

The ordered pair (4, 3) is a solution of the equation. What does the solution (4, 3) represent?

A. Ms. Monti bought 4 adult tickets and 3 children’s tickets.
B. Adult tickets cost $4 each and children’s tickets cost $3 each.
C. Ms. Monti spent $4 on adult tickets and $3 on children’s tickets.
D. The cost of 4 adult tickets equals the cost of 3 children’s tickets.


The growth of a kitten is described by the equation y = 2.5x + 4, where y represents the kitten’s weight in ounces x weeks after it was born. What is the meaning of the fact that the point (4, 14) lies on the graph of the equation?

A. The kitten had an initial weight of 4 ounces.
B. The kitten is growing at a rate of 4 ounces per week.
C. The kitten weighed 4 ounces when it was 14 weeks old.
D. The kitten weighed 14 ounces when it was 4 weeks old.

Answer 1

Answer:

Part 1) Option D $75=10d+5n$

Part 2) Option B $(4, -2), (0, -1), (-4, 0)$

Part 3) Option A $50=2LW$

Part 4) Option A Ms. Monti bought $4$ adult tickets and $3$ children’s tickets

Part 5) Option D The kitten weighed $14$ ounces when it was $4$ weeks old

Step-by-step explanation:

Part 1)

we know that

$1\ dime=\ 0.10$

$1\ nickel=\ 0.05$

Let

d-------> the number of dimes

n-----> the number of nickels

so

$0.10d+0.05n=0.75$

Multiply by $100$ both sides

$10d+5n=75$

Part 2)

we have

$y=(-\frac{1}{4})x-1$

To calculate the solutions of the equation for an x value, substitute the value of x in the equation and find the value of y

For $x=4$ -----> $y=(-\frac{1}{4})(4)-1=-2$ ----> $(4,-2)$

For $x=0$ -----> $y=(-\frac{1}{4})(0)-1=-1$ ----> $(0,-1)$

For $x=-4$ -----> $y=(-\frac{1}{-4})(4)-1=0$ ----> $(-4,0)$

Part 3)

we know that

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$P=50\ ft$

substitute

$50=2(L+W)$ ------> $50/2=(L+W)$ -----> $50=2L+2W$

Par 4)

Let

x------> the number of adult tickets

y------> the number of children tickets

we have

$25x + 15y = 145$

The ordered pair $(4,3)$ represent

$x=4\ adult\ tickets$

$y=3\ children\ tickets$

therefore

Ms. Monti bought $4$ adult tickets and $3$ children’s tickets

Part 5)  

Let

x------> the number of weeks after it was born

y-----> the kitten’s weight in ounces

we have

$y=2.5x+4$  

The ordered pair $(4,14)$ represent

$x=4\ weeks$

$y=14\ ounces$

therefore

The kitten weighed $14$ ounces when it was $4$ weeks old