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3 years ago

Circle R has equation

Mathematics

1 answer:

Rudiy273 years ago

5 0

Answer:

The center is (-10,10) and the radius is 4sqrt(3)

Step-by-step explanation:

(x + 10)^2 + (y - 10)^2 = 48

We can write the equation of a circle as

(x -h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius

(x- -10)^2 + (y - 10)^2 = (sqrt(16*3) )^2

(x- -10)^2 + (y - 10)^2 = (4sqrt(3)) ^2

The center is (-10,10) and the radius is 4sqrt(3)

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Answer:

(2x +30) and (3x -5) (they're opposite to eachother, so they're equal)

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Step-by-step explanation:

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Step-by-step explanation:

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And the cash price is 15,999

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3 years ago

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$ .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$ .....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

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The solution set: $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

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3 years ago

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