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Ahat [919]
2 years ago
15

A teacher is buying supplies for an upcoming art project. She buys x bottles of paint for $1.29 each. She buys 8 more canvases t

han bottles paint for $1.89 each. Her total pretax cost is $37.38. How many of each item is she buying?
Mathematics
2 answers:
Oksi-84 [34.3K]2 years ago
7 0

Answer:

7 bottles of paint and 15 canvases

Step-by-step explanation:

I'm letting x stand for the number of bottles of paint and y stand for the number of canvases

1.29x+1.89y=37.38

and y=x+8

Since the y value is isolated I can just substitute the right side of the y= equation into the original like so

1.29x+1.89(x+8)=37.38

I then distributed the 1.89 to each of the terms in the parentheses

1.29x+1.89x+15.12=37.38

I then combined like terms to get

3.18x+15.12=37.38

Then you subtract 15.12 from both sides to get

3.18x=22.26

and last we divide both sides by 3.18 to get

x=7

Now that we have x, we still need to solve for y, so I take the original equation y=x+8 and substitute in the value we solved for x to get

y=7+8

y=15

Don't forget the labels you gave each variable earlier in the problem when answering!

Elena L [17]2 years ago
4 0

Answer:

She is buying 7 bottles and 15 canvases of paint

Step-by-step explanation:

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3 years ago
Pls show work! questions on pic
Lena [83]

Answer:

a.(10,23)

b.(7,30)

e.(20,20)

Step-by-step explanation:

We are given that

x=Width of garden

y=Length of garden

According to question

x\geq 5

y\geq 20

Perimeter of rectangle=2(x+y)

Where x=Width  of rectangle

y=Length of rectangle

Fencing used =2(x+y)

2(x+y)\leq 80

First we change inequality  into equality

x=5...(1)

y=20...(2)

2(x+y)=80

x+y=40...(3)

Substitute x=0 in equation (3)

y=40

Substitute y=0  in equation (3)

x=40

Substitute x=0 and y=0 in equation (3)

2(x+y)\leq 80

0\leq 80

It is true. Therefore, the  shaded region below the line.

0\geq 5

0\geq 20

These are false equation .Therefore, the shaded region above the line.

Substitute x=5 in equation (3)

5+y=40

y=40-5=35

Substitute y=20 in equation (3)

x+20=40

x=40-20=20

Intersection point of equation (1) and (3) is (5,35) and intersect point of equation (2) and (3) is (20,20).

Now,

a.(10,23)

Substitute in

2(x+y)\leq 80

2(10+23)\leq 80

66\geq 80

10>5 and 23>20

it satisfied all inequality.

Hence, it is a solution.

b.(7,30)

7>5

30>20

2(7+30)=74

it satisfied all inequality.

Hence, it is a solution.

c.(18,25)

18>5

25>20

2(18+25)=86>80

It does not satisfied third inequality

Hence, it is not a solution.

d.(8,35)

8>5

35>20

2(8+35)=86>80

It does not satisfied third inequality

Hence, it is not a solution.

e.(20,20)

2(20+20)=80

20>5

20=20

it satisfied all inequality.

Hence, it is a solution.

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