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marysya [2.9K]
3 years ago
7

018 10.0 points

Mathematics
1 answer:
Elina [12.6K]3 years ago
7 0

Answer:

what grade

Step-by-step explanation:

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The ordered pairs (3,2) and (-2,13) are points on a linear function. Which equation best describes this function?
Mars2501 [29]

Answer:

The equation which describe this function is 5x + 11y = 37

Step-by-step explanation:

<u>To find the slope</u>

slope = (y₂ - y₁)/(x₂ - x₁)

slope = (13 - 2)/(-2 - 3) = 11/-5

<u>To find the equation</u>

(x - 3)/(y - 2) = -11/5

5(x - 3) = -11(y - 2)

5x -15  = -11y + 22

5x + 11y = 22 + 15

5x + 11y = 37

Therefore the equation which describe this function is

5x + 11y = 37

7 0
3 years ago
Read 2 more answers
Ariana bought 7 muffins she paid 13.02 how much did each muffin cost
nalin [4]

Answer:

About $1.86

Step-by-step explanation:

It could be rounded to $1.90.

5 0
3 years ago
Read 2 more answers
A school club sold 300 shirts. 31% were sold to fifth graders, 52% were sold to sixth graders, and the rest were sold to teacher
Yuri [45]
You would do 300• 31/100 to get 90. I’m pretty sure that’s right
4 0
2 years ago
A sum of two mixed numbers is given.<br><br> What is the missing number?
kaheart [24]

Answer:

16.

Step-by-step explanation:

The whole numbers stay the same so it stays as 2. The way they get 30 for the denominator is by multiplying the fraction 1/3 by 10 and 1/5 by 6. So now we have 10/30 and 6/30. Add them together, 10+6=16

4 0
2 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
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