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grigory [225]
4 years ago
6

Find the equation of the perpendicular bisector of the segment AB, if A(1, –2.5) and B(5, 5.5). If the perpendicular bisector of

AB intercepts the y-axis at point P, what are the lengths of PA and PB?
Mathematics
1 answer:
ruslelena [56]4 years ago
5 0
EXPLANATION: First, solve the equation segment for AB using the midpoint formula
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Which of the following is the solution to: -3x + 8 < 14
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Answer:

The correct solution is C (x> -2)

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2 years ago
What is the slope of (4,1) and (6,-1)
schepotkina [342]

Answer:

slope = -1

Step-by-step explanation:

you <u>subtract the y-values</u> (-1-1=-2) then <u>subtract the x-values</u> (6-4=2) then divide the sum of the y-values by the sum of the x-values

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2 years ago
A plane ride lasted 6.4 hours. how many seconds did the plane ride last
ladessa [460]

1 * 60 * 60 = 3600 (seconds in 1 hour)

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8 0
4 years ago
Given the function f(x)=x^2+2x+1, find each value of the function
Paha777 [63]

Answer:

1 9/16

Step-by-step explanation:

F(1/4) means u plug in 1/4 for x

so the equation would be

f(1/4)=1/4^2+2 1/4 + 1

f(1/4)=1/16+1/2+1

f(1/4)= 9/16 +1

f(1/4)= 1 9/16

its not nine teen its one and 9 over 16 just clarification

5 0
3 years ago
Derive the equation of the parabola with the focus <br> is (-7,5) and the directrix of y=-11
Volgvan
So, notice, the focus point is at -7, 5, and the directrix is at y = -11.

keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.

with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.

since the parabola is opening upwards, "p" is positive 8.

\bf \textit{parabola vertex form with focus point distance}\\\\&#10;\begin{array}{llll}&#10;4p(x- h)=(y- k)^2&#10;\\\\&#10;\boxed{4p(y- k)=(x- h)^2}&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ( h, k)\\\\&#10; p=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-------------------------------\\\\&#10;\begin{cases}&#10;h=-7\\&#10;k=-3\\&#10;p=8&#10;\end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2&#10;\\\\\\&#10;32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2&#10;\\\\\\&#10;y=\cfrac{1}{32}(x+7)^2-3

8 0
3 years ago
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