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2 years ago

9

-10(2x+1/5) in standard form

1 answer:

Svetach [21]2 years ago

5 0

Answer:

-20x-2

Step-by-step explanation:

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Y’all someone help pls 5x-3=21-x

attashe74 [19]

5x-3=21-x
6x=24
X=4

You simply have to isolate the variable and then divide by 6 so that x is completely alone.

7 0

3 years ago

Read 2 more answers

Identify the underlined place in 83.5851. Then round the number to that place. The 8 to the right is underlined.

Marina86 [1]

We have to round the number 83.5851 to the nearest hundredth.
If the next smallest place is greater than or equal to 5 we increase the value of the digit we are rounding to by one.
83.5851 ≈ 83.59
Answer: 83.59

8 0

3 years ago

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How do i solve using the rational root theorem. 2x3-9x2-11x+8=0

Lapatulllka [165]

2(3)-9(2)-11x+8=0
(2)(3)-(9)(2)-11x+8=0
6+-18+-11x+8=0
(-11x)+(6+-18+8)=0 Combine Like Terms.
-11x+-4=0
-11x-4=0
Add 4 to both sides.
-11x-4+4=0+4
-11x=4
Divide both sides by -11.
-11x/-11=4/-11
x=-4/11

Hope this helps!

4 0

3 years ago

4. Seleccione la respuesta correcta. Select the right answi

WITCHER [35]

Answer:

B

Step-by-step explanation:

The answer is B I have seen this question befor

5 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

4 years ago