Answer:
149 F
Step-by-step explanation:
If you take -35 and find the absoulte value you get 35 and then add that too 114 you get 149.
Answer:
See explanation and hopefully it answers your question.
Basically because the expression has a hole at x=3.
Step-by-step explanation:
Let h(x)=( x^2-k ) / ( hx-15 )
This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.
Solving for x in that equation:
Adding 15 on both sides:
hx=15
Dividing both sides by h:
x=15/h
For it be a hole, you also must have the numerator is zero at x=15/h.
x^2-k=0 at x=15/h gives:
(15/h)^2-k=0
225/h^2-k=0
k=225/h^2
So if we wanted to evaluate the following limit:
Lim x->15/h ( x^2-k ) / ( hx-15 )
Or
Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.
We were ask to evaluate
Lim x->3 ( x^2-k ) / ( hx-15 )
Comparing the two limits h=5 and k=225/h^2=225/25=9.
Answer:
x = 16
Explanations:
Since △ACD ~ △ABE, it means that:
CD/AD = BE/AE..........(*)
CD = 3x + 8
AD = AE + ED
AD = (x - 1) + 27
AD = x + 26
BE = 20
AE = x - 1
Substituting CD, AD, BE, and AE into equation (*)
(3x+8)/(x+26) = 20/x-1
Cross multiply:
(3x+8)(x-1) = 20(x+26)
3x² - 3x + 8x - 8= 20x + 520
3x²- 15x - 528 = 0
Divide through by 3
x² - 5x - 176 = 0
Solving the quadratic equation above for the values of x
x² - 16x + 11x - 176 = 0
x(x - 16) + 11 (x - 16) = 0
(x - 16)(x + 11) = 0
If x - 16 = 0
x = 16
If x + 11 = 0
x = -11
x cannot be -11 because x = -11 will make side CD and AE to be negative, and the sides of a triangle cannot be negative.
Therefore, x = 16
Answer: D. 2x² + 3/2.x - 5
Step-by-step explanation:
f(x) - n - move the graph n units down
f(x) + n - move the graph n units up
f(x - n) - move the graph n units right
f(x + n) - move the graph n units left
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![f(x)=x^2\\\\g(x)=f(x)-3=x^2-3](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2%5C%5C%5C%5Cg%28x%29%3Df%28x%29-3%3Dx%5E2-3)
<h3>Answer: g(x) = x² - 3</h3>