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Furkat [3]
3 years ago
11

ASAP! WILL MARK BRAINLIEST!!

Mathematics
2 answers:
Naya [18.7K]3 years ago
5 0

Answer:

y=x+5+x+5?

Step-by-step explanation:

I'm not sure and I did this not that long ago

Anton [14]3 years ago
5 0

Answer:

10

Step-by-step explanation:

y=|x-5|+|x+5|

= 5-x+x+5

=10

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I need to find the product help
Licemer1 [7]
I'm assuming you mean

2\times\frac{17}{10}

Write the 2 as a fraction 2/1 and multiply.

\frac{2}{1}\times\frac{17}{10}=\frac{34}{10}=\frac{17}{5}


7 0
3 years ago
Simplify the expression cos x cot x+ sin x please select the best answer from the choices provided a.0, b.csc x, c. Tan x, d sec
expeople1 [14]

Answer:

cot x = \frac{cos x}{sin x}

cos x \frac{cos x}{sin x} + sin x

\frac{cos^2 x}{sin x} +sin x

sin^2 x + cos^2 x =1

Solving for cos^2 x we got cos^2 x =1 -sin^2 x and replacing this we got:

\frac{1-sin^2 x}{sin x} +sin x

\frac{1}{sin x} -\frac{sin^2 x}{sin x} +sin x

csc x -sin x + sin x = csc x

And then the best option for this case would be:

b.csc x

Step-by-step explanation:

For this case we have the following expression given:

cos x cot x + sin x

We know from math properties that the definition for cot is cot x = \frac{cos x}{sin x}

If we use this definition we got:

cos x \frac{cos x}{sin x} + sin x

\frac{cos^2 x}{sin x} +sin x

Now we can use the following identity:

sin^2 x + cos^2 x =1

Solving for cos^2 x we got cos^2 x =1 -sin^2 x and replacing this we got:

\frac{1-sin^2 x}{sin x} +sin x

\frac{1}{sin x} -\frac{sin^2 x}{sin x} +sin x

csc x -sin x + sin x = csc x

And then the best option for this case would be:

b.csc x

4 0
3 years ago
Complete the scale by labeling the remaining tick marks​
scoundrel [369]

Answer:

Can you provide a picture?

I cannot help you if you don't.

5 0
2 years ago
25% of what number is 21
chubhunter [2.5K]
If we call the number "x" then:
x*25%=21
x*25/100=21
x=21*4
x=84
8 0
3 years ago
How do you estimate a whole number 3,576
frozen [14]
By dividing it by a number that = to it
5 0
3 years ago
Read 2 more answers
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