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Delvig [45]
3 years ago
6

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

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Step-by-step explanation:

I hope this helps :))

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3 years ago
An appliance is for sale at either (a) P15,999 cash or (b) on terms, P1,499 each month for the next 12 months.  Money is 9% comp
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Cash price

Step-by-step explanation:

The computation is shown below:

The Interest rate per month (r) = (9% ÷ 12) = 0.75%

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= 1,499 × {[(1 + 0.75%)^12 - 1] ÷ 0.75%}

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Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

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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}

or

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

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

or

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

Simplify:

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

Substitute y = cos x we have;

\cos x = 1

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and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

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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