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ira [324]
3 years ago
15

Find the equation of a line given the point and slope below. Arrange your answer in the form y = mx + b, where b is the constant

.
(7, 1)
m = 10
Can someone help me and show me how to do these the following lessons are all on this so tyy owo
Mathematics
2 answers:
Nana76 [90]3 years ago
5 0

Answer:

Step-by-step explanation:

an equation is :  y = 10x+b

calculate b : this line passes by : (7;1) when x =7  y = 1

so : 10(7)+ b=1

b= -69

so : y = 10x-69

Liula [17]3 years ago
4 0

Answer:

y = 10x - 69.

Step-by-step explanation:

Start with the slope-intercept equation of a straight line, y = mx + b.  We substitute 1 for y, 7 for x, and 10 for m, obtaining:

1 = 10(7) + b

Subtracting 70 from both sides, we get -69 = b.

Thus, the desired equation is y = 10x - 69.

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Given right triangle ABC with a=50, and m
kramer

Answer:

A=50 B=40 C=90

Step-by-step explanation:

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7 0
3 years ago
What is the perpendicular distance from the line y= 2x-3 to the point (8,3)
Liula [17]
Find the perpendicular line then find the intersection then find the point

perpendicular lines have slopes that are perpendicular
the slopes multiply bo -1

y=mx+b
m=slope
y=2x-3
2 is slope

2 times what=-1
what=-1/2


the equation is
y-3=-1/2(x-8) or
y=(-1/2)x+7

find intersection
at (4,5)
distance bwetweeen (8,3) and (4,5)
D=\sqrt{(8-4)^2+(3-5)^2}
D=\sqrt{(4)^2+(-2)^2}
D=\sqrt{16+4}
D=\sqrt{20}
D=2√5
distance= 2√5
7 0
3 years ago
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