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Mathematics 20 Online
OpenStudy (anonymous):

find the value of the varibles x=e,5,7,10,12 y=10,13,19,f,34 and explain how you got i will MEDAL AND FAN who ever helps me !!!

OpenStudy (anonymous):

@Kstate

OpenStudy (anonymous):

@danid16

OpenStudy (anonymous):

@rational

jimthompson5910 (jim_thompson5910):

from the given values, we see that (5,13) and (7,19) are two points on this graph agreed?

OpenStudy (anonymous):

i know what they are i just dont know why they are that answer

jimthompson5910 (jim_thompson5910):

sorry I'm just seeing this message

jimthompson5910 (jim_thompson5910):

using (5,13) and (7,19), let's find the slope m = (y2 - y1)/(x2 - x1) m = (19-13)/(7-5) m = 6/2 m = 3 So the slope of this line is m = 3

jimthompson5910 (jim_thompson5910):

hopefully you can see how I got that

OpenStudy (anonymous):

iam not finding slope

jimthompson5910 (jim_thompson5910):

we will use the slope to find the equation of the line

jimthompson5910 (jim_thompson5910):

once we know the equation of the line, we can use that to find e and f

OpenStudy (anonymous):

f=28 and e=4 i just dont know how

jimthompson5910 (jim_thompson5910):

let me continue and I'll show you do you see how I got that slope?

OpenStudy (anonymous):

yea

jimthompson5910 (jim_thompson5910):

we know that (5,13) is a point on the line, so x = 5 and y = 13 m = 3 is the slope

jimthompson5910 (jim_thompson5910):

y = mx+b y = 3x + b ... plug in the slope 13 = 3(5) + b ... plug in the point (5,13) 13 = 15 + b 13 - 15 = b -2 = b b = -2 So we know that m = 3 and b = -2. Therefore the equation is y = 3x - 2

jimthompson5910 (jim_thompson5910):

Now to find the value of e y = 3x - 2 10 = 3x - 2 ... plug in y = 10 and solve for x 10+2 = 3x 3x = 12 x = 12/3 x = 4 So e = 4

jimthompson5910 (jim_thompson5910):

To find the value of f, you plug x = 10 into y = 3x - 2 and evaluate

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

\[y=3x-2\]

jimthompson5910 (jim_thompson5910):

yes now plug in x = 10

OpenStudy (anonymous):

oh ok i got it thhank you

jimthompson5910 (jim_thompson5910):

you're welcome

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