Answer: y=0
Step-by-step explanation: i think this is right
Answer:
The equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4) is x=-5
Option A is correct.
Step-by-step explanation:
We need to find equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4)
We need a line parallel to x=-2 or x+2=0 it should be of form x+k=0
We need to find k, by putting value of x=-5 as given in question the point(-5,4)
-5+k=0
k=5
So, the equation of line will be found by putting k=5:
x+k=0
x+5=0
x=-5
So, the equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4) is x=-5
Option A is correct.
B.
<span>$5852.86
Working;
</span><span>A = P(1 + r)^t
A=5000(1+</span>

)^5
A=<span>5852.86</span>
4/(x+1) = 3/x + 1/15
Should we make common denominators with everything, we get
4*15*x / [15x(x+1)] = 3*15*(x+1)/[15x(x+1)] + x(x+1)/[15x(x+1)]
Multiply both sides of the equation by the denominator to cancel them
60x = 45(x+1) + x(x+1)
60x = 45x + 45 + x^2 + x
x^2 - 14x + 45 = 0
(x-9)(x-5) = 0
The answer to this question is that the solutions are x=9 and x=5.
Answer:
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Step-by-step explanation:
We have the equation of the position of the object
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We need to solve the equation for the variable a

Subtract
and
on both sides of the equality


multiply by 2 on both sides of equality


Divide between
on both sides of the equation

Finally
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