Ok so to start you plug in what’s in the parentheses for x
4c=8-5(4c)
2-k=8-5(2-k)
4p+3=8-5(4p+3)
In the first one you distribute the -5
4c=8-20c
Then add 20c to both sides
24c=8
And divide by 24
C=8/24 or 1/3
Distributing the -5 on the second one gets you
2-k=8-10+5k
Then you combine like terms
2-k=-2+5k
Subtract 5k from both sides
2-6k=-2
Subtract 2 from both sides
-6k=-4
Divide by -6
K=4/6 or 2/3
and lastly the equation is 4p+3=8-5(4p+3)
So you distribute -5
4p+3=8-20p-15
Combine like terms
4p+3=-20p-7
Add 20p to both sides
24p+3=-7
Subtract 3 from both sides
24p=-10
And then divide by 24
P= -10/24 or -5/12
Answer:
t = 5
Step-by-step explanation:
Given
t + 2 = 7 ( isolate t by subtracting 2 from both sides )
t = 7 - 2 = 5
The speed of the ball is
ds/dt = 32t
At t =1/2 s
ds/dt = 16 ft/s
The distance from the ground
50 - 16(1/2)^2 = 46 ft
The triangles formed are similar
50/46 = (30 + x)/x
x = 345 ft
50 / (50 - s) = (30 + x)/x
Taking the derivative and substituing
ds/dt = 16
and
Solve for dx/dt
Answer:
Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line
is
Step-by-step explanation:
Given:
To Find:
Equation of line passing through ( 16, -7) and is perpendicular to the line
Solution:
...........Given
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Comparing with,
Where m =slope
We get
We know that for Perpendicular lines have product slopes = -1.

Substituting m1 we get m2 as
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Therefore the slope of the required line passing through (16 , -7) will have the slope,
Now the equation of line in slope point form given by
Substituting the point (16 , -7) and slope m2 we will get the required equation of the line,
Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line
is