Find the equation of the line with a slope of 3/4 that goes through the point (8,10)
2 answers:
the answer is ......... y=3/4+10
This is good.
Using the general format: y -y1 = m(x-x1), format of a slope and a point.
(x1, y1) = (8,10)
y -10 = (3/4)*(x-8)
y -10 = 3x/4 -6 Multiply through by 4.
4y -40 = 3x -24
4y = 3x -24 + 40
4y = 3x + 16.
So equation of the line is: 4y = 3x + 16.
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11/6 + -2/5 +-13/10?
11/6 + (-2/5) + (-13/10)
43/30 - 13/10
2/15
Answer:
B
Step-by-step explanation:
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P(x)=R(x)-C(x)
=(-0.5x²+800x-100)-(300x+250)
=-0.5x²+800x-100-300x-250
=-0.5x²+800x-300x-100-250
=-0.5x²+500x-350 (2)
Answer:
ㄥBEC=70;ㄥABE=160
Step-by-step explanation:
(x-5)+(3x-5)+90=180
x-5+3x-5+90=180
4x+80=180
4x=100
x=25
ㄥBEC=3*25-5=70
ㄥABE=180-(25-5)=160