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3 years ago

If (−4, 11) and (−6, 5) are the endpoints of a diameter of a circle, what is the equation of the circle?

Mathematics

2 answers:

aliina [53]3 years ago

6 0

Answer:

The anser is B 100% willing to put $100 on it.

Step-by-step explanation:

lora16 [44]3 years ago

4 0

Answer:

A. $(x+5)^2+(y-8)^2=100$

Step-by-step explanation:

1. The standard form of the equation of a circle is $(x-h)^{2} + (y-k)^2=r^2$

2. To find the center we require the midpoint of the 2 given points

center = $[\frac{1}{2}(-4-6),\frac{1}{2}(11+5)]$

=(-5,8)

3. The radius is the distance from the centre to either of the 2 given points calculate the radius using the distance formula

d= $\sqrt{(x_{2}+x_{1})^{2} +(y_{2}-y_{1})^{2}$

let $(x_{1},y_{1})= (-5,8) \\$ and $(x_2,y_2)=(-4,11)$

r= $\sqrt{(-4+5)^2+(11-8)^2} =\sqrt{81+9}=10$

--> $(x-(-5)^2+(y-8)^2=10^2$

--> $(x+5)^2+(y-8)^2=100$

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Answer:

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Step-by-step explanation:

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Pavlova-9 [17]

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3 years ago

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y = -3x + 5

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3 years ago

Solve using the quadratic formula. 2x2=8x-7

Ierofanga [76]

Answer:

$\boxed{x=\frac{4\pm\sqrt{2}}{2}}$

Step-by-step explanation:

Part 1: Rewriting equation to match ax² + bx + c = 0 (quadratic function)

The given equation is not written in quadratic form. To rewrite the equation:

All values need to be on the left side of the equation and set equal to zero.

To overcome this difficulty, follow these mathematical steps:

$2x^2=8x-7\\2x^2-8x=-7\\2x^2-8x+7=0$

Subtract 8x from both sides of the equation to rearrange it to the left side. Then, add 7 to rearrange it as well. Finally, set the three values on the left of the equation equal to zero.

Part 2: Using the quadratic formula

The quadratic formula is defined as $\boxed{x=\frac{-b\pm\sqrt{b^2-4ac} }{2a} }$ .

Using the parent quadratic function, the values are easy to find in the given equation. $\boxed{a=2, b=-8, c=7}$

Substitute these values into the quadratic formula and solve for x.

$x=\frac{8\pm\sqrt{(-8)^2-4(2)(7)}}{2(2)} \\\\x=\frac{8\pm\sqrt{64-4(14)}}{4}\\\\x=\frac{8\pm\sqrt{64-56}}{4} \\\\x=\frac{8\pm\sqrt{8}}{4}\\\\x= 2\pm\frac{\sqrt{8}}{4}\\ \\x=2\pm\frac{\sqrt{2}}{2}$

Part 3: Solving for x with the values from the quadratic formula

Now that x is set equal to the simplified version of the equation, the operations have to be followed through with.

This equation will have two zeros/roots to solve for by setting x equal to zero.

Operation 1: Addition

$x=2+\frac{\sqrt{2} }{2}\\\\x=\frac{4+\sqrt{2}}{2}$

Operation 2: Subtraction

$x=2-\frac{\sqrt{2}}{2}\\ \\x=\frac{4-\sqrt{2}}{2}$

Because both values are the exact same (minus the operations), the roots can be simplified even further to one value:

$\boxed{x=\frac{4\pm\sqrt{2}}{2}}$

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