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scZoUnD [109]
3 years ago
8

Please help! I need to tell whether x and y are in a proportional relationship and I need to write an equation that represents t

he relationship if it’s proportional.

Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

It is a proportional relationship

Step-by-step explanation:

To see if multiple ratios are proportional, you could write them as fractions, reduce them, and compare them. If the reduced fractions are all the same, then you have proportional ratios.The equation that represents a proportional relationship, or a line, is y=kx, where k is the constant of proportionality.

Hope this helps

the constant of proportionality is 2

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Murrr4er [49]
Muliply 65 and 0.12 and get 7.8

Subtract 65-7.8 to get marked down price

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3 years ago
There was an activity planned outdoors, but it started to rain. They took a vote and 45 people of the 60 people wanted to stay i
Elza [17]

Answer:

15/60×100=25%

Step-by-step explanation:

the number of students wanted to go outside/the whole number of students×100

15/60×100=25%

4 0
3 years ago
Find the perimeter of triangle ABC with vertices A(6, 3), B(6, - 2) , and C(- 4, 3) .
anyanavicka [17]

Answer:

26.2 units

Step-by-step explanation:

We are given the points/vertices

A(6, 3),

B(6, - 2) , and

C(- 4, 3)

Step two

Let us find the distances between the given points/vertices

A-B =A(6, 3) to B(6,-2)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6-6)²+(-2-3)²)

d=√(-2-3)²)

d=√(-5)²)

d=5 units

B-C=B(6, - 2) to C(-4, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((-4-6)²+(3+2)²)

d=√(-10)²+(5)²)

d=√100+25

d=√125

d=11.2 units

C-A=C(-4, 3) to A(6, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6+4)²+(3-3)²)

d=√(10)²

d=√100

d=10 units

Hence the perimeter is 5+11.2+10

P=26.2 units

4 0
3 years ago
The point Z(5, -3) is rotated 270° clockwise around the origin. What are the coordinates of the resulting point, Z′?
g100num [7]

Answer:

(3, - 5 )

Step-by-step explanation:

Under a clockwise rotation about the origin of 270°

a point (x, y ) → (- y, x ), hence

Z(5, - 3 ) → Z'(3, - 5 )

4 0
3 years ago
X+y+z=12<br> 6x-2y+z=16<br> 3x+4y+2z=28<br> What does x, y, and z equal?
lianna [129]

Answer:

x = 20/13 , y = 16/13 , z = 120/13

Step-by-step explanation:

Solve the following system:

{x + y + z = 12 | (equation 1)

6 x - 2 y + z = 16 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Swap equation 1 with equation 2:

{6 x - 2 y + z = 16 | (equation 1)

x + y + z = 12 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/6 × (equation 1) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+(4 y)/3 + (5 z)/6 = 28/3 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Multiply equation 2 by 6:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/2 × (equation 1) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+5 y + (3 z)/2 = 20 | (equation 3)

Multiply equation 3 by 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+10 y + 3 z = 40 | (equation 3)

Swap equation 2 with equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+8 y + 5 z = 56 | (equation 3)

Subtract 4/5 × (equation 2) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+(13 z)/5 = 24 | (equation 3)

Multiply equation 3 by 5:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+13 z = 120 | (equation 3)

Divide equation 3 by 13:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract 3 × (equation 3) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y+0 z = 160/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 2 by 10:

{6 x - 2 y + z = 16 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Add 2 × (equation 2) to equation 1:

{6 x + 0 y+z = 240/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract equation 3 from equation 1:

{6 x+0 y+0 z = 120/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 1 by 6:

{x+0 y+0 z = 20/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Collect results:

Answer:  {x = 20/13 , y = 16/13 , z = 120/13

8 0
3 years ago
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