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4 years ago

What’s the correct answer for this?

Mathematics

2 answers:

Ierofanga [76]4 years ago

6 0

Answer:

The explanation below should guide you to solve it's

Step-by-step explanation:

x2 + y2 - 14x -18y +105 =0

Now the standard form a the equation of a circle is

(X-Xo)2 + (y-yo)2 = r2

The above expressions becomes

x2- 14x + (-14/2)2 - (- 14/2)2 + y2 -18y + (-18/2)2 - (-18/2)2 + 105 =0

The above manipulation does not alter the equation and it's a way of forming squares of a quadratic equation.

(x - 7)2 + (y - 9)2 -49 - 81 + 105 = 0

(x - 7)2 + (y - 9)2 -130 + 105 = 0

(x - 7)2 + (y - 9)2 -25 = 0

(x - 7)2 + (y - 9)2= 25

From the analysis above you can determine the centres of the circle denoted by Xo and Yo as well as the radius which is the square root of the expression at the right

Lunna [17]4 years ago

5 0

Answer:

Step-by-step explanation:

In the attached file

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3 years ago

Right triangle XYZ has right angle Z. If the sin(X)=1213 , what is the cos(X)

AlekseyPX

Given:

Right triangle XYZ has right angle Z.

$\sin(x)=\dfrac{12}{13}$

To find:

The value of $\cos x$ .

Solution:

We know that,

$\sin^2(x)+\cos^2(x)=1$

$\cos^2(x)=1-\sin^2(x)$

$\cos(x)=\pm\sqrt{1-\sin^2x}$

For a triangle, all trigonometric ratios are positive. So,

$\cos(x)=\sqrt{1-\sin^2x}$

It is given that $\sin(x)=\dfrac{12}{13}$ . After substituting this value in the above equation, we get

$\cos(x)=\sqrt{1-(\dfrac{12}{13})^2}$

$\cos(x)=\sqrt{1-\dfrac{144}{169}}$

$\cos(x)=\sqrt{\dfrac{169-144}{169}}$

$\cos(x)=\sqrt{\dfrac{25}{169}}$

On further simplification, we get

$\cos(x)=\dfrac{\sqrt{25}}{\sqrt{169}}$

$\cos(x)=\dfrac{5}{13}$

Therefore, the required value is $\cos(x)=\dfrac{5}{13}$ .

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Graph on a number line. −6 over 7

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