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3 years ago

Which point lies on the circle centered at the origin with radius V5?

Mathematics

1 answer:

kiruha [24]3 years ago

8 0

Answer:

Work it out

Step-by-step explanation:

1. Work it out

2.Work it out!!!!

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Find the exact value of cos(a+b) if cos a=-1/3 and cos b=-1/4 if the terminal side if a lies in quadrant 3 and the terminal side

maria [59]

Answer:

cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = $-\frac{1}{3}$

cos(b) = $-\frac{1}{4}$

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = $-\sqrt{1-(-\frac{1}{3})^2}$ [Since, sin(a) = $\sqrt{(1-\text{cos}^2a)}$ ]

= $-\sqrt{\frac{8}{9}}$

= $-\frac{2\sqrt{2}}{3}$

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be negative.

sin(b) = $-\sqrt{1-(-\frac{1}{4})^2}$

= $-\sqrt{\frac{15}{16}}$

= $-\frac{\sqrt{15}}{4}$

By substituting these values in the identity,

cos(a + b) = $(-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})$

= $\frac{1}{12}-\frac{\sqrt{120}}{12}$

= $\frac{1}{12}(1-\sqrt{120})$

= $\frac{1}{12}(1-2\sqrt{30})$

Therefore, cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

5 0

3 years ago

0.01 written as a percent then round to the nearest tenth of a percent

Aleks [24]

Answer and step-by-step explanation:

To write it as percent, multiply it with 100, we get:

0.01 * 100 = 1%

Nearest tenth of a percent would be 10%, 20%, 30%, etc.

So as we can see 1% is nearest to 10% so we round it up to 10%

Hope this helped :3

7 0

3 years ago

Write the total money amount.then write the amount as a fraction or mixed number and as a decimal in terms of dollars

Kitty [74]

Hhczfik in nb vguiigdr 5th hb. jigfddtjnbvt

6 0

3 years ago

What are the solutions to this equation?

ziro4ka [17]

$\displaystyle\\(x-4)^2=49\\\\(x-4)^2-49=0\\\\(x-4)^2-7^2=0\\\\(x-4-7)(x-4+7)=0\\\\(x-11)(x+3)=0\\\\x-11 = 0~~~\text{or}~~~x+3=0\\\\x-11=0~~~\implies~~~\boxed{x_1=11}\\\\x+3=0~~~\implies~~~\boxed{x_2=-3}\\\\\boxed{\bf The~solutions~are:~~x =11~~\text{or}~~x =-3}$

8 0

3 years ago

How to solve X^4-19x^2+48=0

ZanzabumX [31]

Two numbers that add up to -19 and multiply to 48 are -16 and -3:

$(x^4-19x^2+48)=0\\(x^2-16)(x^2-3)=0\\(x+4)(x-4)(x^2-3)=0$

So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.

x+4=0
x = -4

x-4=0
x = 4

x^2-3=0
x^2 = 3
x = +/- √3

So, the solutions are -4, -√3, √3, and 4.

4 0

3 years ago