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hodyreva [135]
2 years ago
13

What is the vertex of the parabola defined by the equation

Mathematics
1 answer:
avanturin [10]2 years ago
7 0

Answer: B. (2,2)

Step-by-step explanation:

The vertex of the parabola in the form (x-h)^2 = 4p(y-k) is (h, k).

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Write an equation in slope-intercept form of<br> the line that passes through (-6,4) and<br> (-1,2).
xz_007 [3.2K]

Answer:

Step-by-step explanation:

(-6 , 4)  & (-1 , 2)

Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

         = \frac{2-4}{-1-[-6]}\\\\= \frac{-2}{-1+6}\\\\= \frac{-2}{5}

m = -2/5   & (-6 , 4)

y -y1 = m(x -x1)

y - 4 = \frac{-2}{5}(x - [-6])\\\\y - 4 = \frac{-2}{5}(x + 6)\\\\y - 4 = \frac{-2}{5}x + 6*\frac{-2}{5}\\\\y = \frac{-2}{5}x -\frac{12}{5}+4\\\\y=\frac{-2}{5}x-\frac{12}{5}+\frac{4*5}{1*5}\\\\y=\frac{-2}{5}x-\frac{12}{5}+\frac{20}{5}\\\\y=\frac{-2}{5}x+\frac{8}{5}

6 0
3 years ago
What is the volume of the sphere shown below?
Angelina_Jolie [31]
The answer is b !
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8 0
3 years ago
Solve 11b^2-9=-68 using the square root method
Vladimir79 [104]

Answer:

b=i*\sqrt{\frac{59}{11} } or -i*\sqrt{\frac{59}{11} }

Step-by-step explanation:

11b^2-9=-68

11b^2=-59

b^2=-59/11

b=i*\sqrt{\frac{59}{11} } or -i*\sqrt{\frac{59}{11} }

"i" in this case is an imaginary number, equal to \sqrt{-1\\}

if you haven't learned about these yet, something is wrong with the question

3 0
3 years ago
1.What is the solution to the system of equations?
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1. x=4, y=0
3. x=2.6666, y=0.3333
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