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2 years ago

Help please ASAP ASAP

Mathematics

1 answer:

AveGali [126]2 years ago

3 0

Answer:

Step-by-step explanation:

At the y axis the number is 4

At the x axis the number is -5

So 4-(-5)=9

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Answer please and thank you

USPshnik [31]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Tom earns $32,110 annually. What is the total amount of

sveticcg [70]

Bi weekly means every two weeks, divide 12 (this would be your months) by your annual pay (32,110), and you would get $2675.83. I hope this helps some.

8 0

3 years ago

A number squared increased by 5, is 41

Ksenya-84 [330]

Answer:

n² + 5 = 41 n = 6

Step-by-step explanation:

n² + 5 = 41

n² + 5 - 5 = 41 - 5

n² = 36

What squared equals 36? 6!

6² = 6 * 6 = 36

Check your answer!

6² + 5 = 41

36 + 5 = 41

41 = 41

Have a good night! :)

7 0

3 years ago

Help! Prove the equality arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

Proof in the explanation

Step-by-step explanation:

Trigonometric Equalities

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

5 0

3 years ago

Factor completely, 0.2x + 0.8 a. x b. 0.8(x + 1) c. 0.2(x + 1) d. 0.2(x + 4)

Zepler [3.9K]

Answer:

d. 0.2 ( x + 4 )

Step-by-step explanation:

0.2x + 0.8

Both 0.2x and 0.8 have a common factor of 0.2, hence pull 0.2 out.

0.2x + 0.8

0.2 ( x + 4 )

5 0

3 years ago