Find the equation of a line which passes through the point (3, -1) with the gradient M=2÷3
1 answer:
Step-by-step explanation:
Equ of line -
y=mx+c
Using point (3,-1)
3 as x, - 1 as y
M=2/3
Substitute the values of these coordinates
3=2/3(-1) +c - - - - now find for c
c= 3- (-2/3)
c= 3.67
Now with the value of c, put the values back into the equation form .. Assuming we don't have a point to work with
y=2/3x + 3.67
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Answer:
Y-int (0,-6)
x-int (-8,0)
Step-by-step explanation:
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Answer:
x = 32
Step-by-step explanation:
∠BCA = ∠DBA (90 - ∠DBC)
∠A = ∠A
ΔABD similar to ΔACB
AC/AB = AB/AD
x / 8 = 8 / 2
x = 32
Answer: 12
Step-by-step explanation:
Given:
A = 452
r = √A/3
Step 1: Substitute 452sqf for A
r = √452/3
Step 2: Solve
452 ÷ 3 = 150.66∞
√150.66∞ = 12.27∞
12.27∞ rounded to the nearest integer equal 12
