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3 years ago

What is the absolute value of -6

Mathematics

2 answers:

patriot [66]3 years ago

8 0

Answer:

Step-by-step explanation:

Absolute value just means you make negatives into positives so it's just 6.

valkas [14]3 years ago

7 0

Answer:

Step-by-step explanation:

When it comes to absolute value is the opposite of the negative number but if the number is positive it stays the same.

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Elana walked 12 miles. then she walked 1/4 that distance. how far did she walk all together. select all that apply.

Nezavi [6.7K]

Answer:

c., d., f

Step-by-step explanation:

"elana walked 12 miles"

That is 12.

"then she walked 1/4 that distance"

That is 1/4 of 12 miles, so it is 1/4 * 12

She walked altogether

12 + 1/4 * 12 = 12 + 3 = 15

She walked altogether 15 miles.

a.12+1/4 = 12.25 which is not 15

b.12×1/4 = 3 which is not 15

c.12+1/4×12 = 12 + 3 = 15 which is correct

d.12(1+ 1/4) = 12(1.25) = 15 which is correct

e.12×3/4 = 9 which is not 15

f.12×5/4 = 15 which is correct

Answer: c., d., f

3 0

3 years ago

Find the value of 151 - 4(32 - 2).<br> -11<br> -33<br> 7<br> -23

egoroff_w [7]

Steps to solve:

151 - 4(32 - 2)

~Distribute

151 - 128 - 8

~Subtract

23 - 8

Best of Luck!

3 0

3 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

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