The vertex of the graph of y=(x-1)^2-5 is

Answers

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!

step-by-step explanation:

12475909

For y  = (x-h)^2 + k
the vertex is h,k

comparing
y=(x-1)^2-5  with  y  = (x-h)^2 + k

we have h = 1, k = -5

so the vertex of y=(x-1)^2-5  is  (1,-5)


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