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Dafna11 [192]
2 years ago
5

Find all real values of $x$ such that $x^2-5x + 4=0$.

Mathematics
2 answers:
IRISSAK [1]2 years ago
8 0

Answer:

Find the 58th term of the arithmetic sequence

−20

Step-by-step explanation:

Luda [366]2 years ago
5 0

Answer:

x=4 && x=1

Step-by-step explanation:

x^2 - x - 4x + 4=0

x(x-1) - 4(x - 1)=0

(x-4)(x-1)=0

X=4 and x=1

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Need help with this question!!!<br> Pls explain!!!<br><br> 3=-(4x+7)+3x
natka813 [3]

The answer would be, x = 10

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3 years ago
Read 2 more answers
Line segment XY has endpoints X(5,7) and Y(-3,3)
dem82 [27]

y=-2x+7

Step-by-step explanation:

line segment xy has endpoints x(5 7) and y(-3 3)

for the equation of the perpendicular bisector of line segment xy

slope of line segment xy = 7-3 / 5+3

= 4/8

=1/2

so slope of perpendicular bisector is -2

as m1m2= -1 or m1= -1/m2

As perpendicular bisector goes through midpoint of xy , let's find midpoint of xy

midpoint(x,y) = (5-3 / 2 , 7+3 / 2)

=(2/2, 10/2)

=(1,5)

find the equation of line(perpendicular bisector) passing through (1,5) and the slope -2

y-5 = -2(x-1)

y-5 =-2x+2

y=-2x+7

6 0
3 years ago
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