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erica [24]
3 years ago
7

Write the equation of the parabola that has the vertex at point (1,3) and passes through the point (−1,5).

Mathematics
1 answer:
Andrei [34K]3 years ago
6 0

Answer:

y = \frac{1}{2} x² - x + \frac{7}{2}

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (1, 3 ), thus

y = a(x - 1)² + 3

To find a substitute (- 1, 5) into the equation

5 = a(- 1 - 1)² + 3 ( subtract 3 from both sides )

2 = 4a ( divide both sides by 4 )

a = \frac{1}{2} , thus

y = \frac{1}{2} (x - 1)² + 3 ← expand factor using FOIL

y = \frac{1}{2} (x² - 2x + 1) + 3 ← distribute parenthesis

  = \frac{1}{2} x² - x + \frac{1}{2} + 3

  = \frac{1}{2} x² - x + \frac{7}{2}

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If

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Sarah needs 30 liters of a 25% acid solution how many liters of the 10% and the 30% acid solutions should she mix to get what sh
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7.5 L of 10% solution and 22.5 L of 30% solution

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Volume of 10% solution plus volume of 30% solution = total volume of 25% volume.

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Solve the system of equations, using either substitution or elimination.  I'll use substitution:

x = 30 − y

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3 − 0.10 y + 0.30 y = 7.5

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y = 22.5

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Sarah needs 7.5 L of 10% solution and 22.5 L of 30% solution.

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Identify the center and radius from the equation of the circle given below. x^2+y^2+121-20y=-10x
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Answer:

Center: (-5,10)

Radius: 2

Step-by-step explanation:

The equation of the circle in center-radius form is:

(x-h)^2+(y-k)^2=r^2

Where the point (h,k)  is the center of the circle and "r" is the radius.

Subtract 121 from both sides of the equation:

x^2+y^2+121-20y-121=-10x-121\\x^2+y^2-20y=-10x-121

Add 10x to both sides:

x^2+y^2-20y+10x=-10x-121+10x\\x^2+y^2-20y+10x=-121

Make two groups for variable "x" and variable "y":

(x^2+10x)+(y^2-20y)=-121

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Add (\frac{10}{2})^2=5^2 inside the parentheses of "x".

Add  (\frac{20}{2})^2=10^2  inside the parentheses of "y".

Add 5^2 and 10^2 to the right side of the equation.

Then:

(x^2+10x+5^2)+(y^2-20y+10^2)=-121+5^2+10^2\\(x^2+10x+5^2)+(y^2-20y+10^2)=4

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You can observe that the radius of the circle is:

r=2

And the center is:

(h,k)=(-5,10)

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