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2 years ago

15

I need help please help

2 answers:

sveticcg [70]2 years ago

5 0

Answer:

D

Step-by-step explanation:

Artist 52 [7]2 years ago

3 0

The answer is d, since the area is 74 it’ll equal to 74, u have to multiply not add as u did

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Maksim231197 [3]

Answer:

D. x=1

Step-by-step explanation:

Image can be rewritten as 3x + 6 = 9.

Solve for x.

3x +6 = 9

Isolate x by subtracting 6 from both sides.

3x = 3

Get rid of the 3 by dividing by 3 on both sides.

x = 1

Hope this helped.

8 0

2 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

2 years ago

Please someone help me to prove this.

dem82 [27]

<h3><u>Answer</u> :</h3>

We know that,

$\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}$

$\dag\bf\:sin2A=2sinA\:cosA$

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

$\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}$

$\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}$

$\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}$

$\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}$

$\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}$

$\leadsto\:\bf{tan(A+B)}$

5 0

2 years ago

Read 2 more answers

Determine whether the given value of the variable is the solution of the equation (4-9)

Katen [24]

Answer: what do you think is the answer

Step-by-step explanation:

6 0

2 years ago

I need help what is this pemdas problem (6*3/7*7) 2+2/8

mart [117]

Gimme a second ill help ya

7 0

2 years ago

Read 2 more answers