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3 years ago

HELP PLEASE THIS IS DUE IN 30 MIN PLEASE HELP URGENT

Mathematics

2 answers:

DanielleElmas [232]3 years ago

5 0

What the other guy said

Oxana [17]3 years ago

3 0

Answer:

12.5

Step-by-step explanation:

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Solve the following inequality.45 < 9(x + 3) < 153

oksian1 [2.3K]

In order to solve this inequality, we can do the following steps:

[tex]\begin{gathered} 45<9(x+3)<153\\ \\ \frac{45}{9}Therefore the correct option is B.

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1 year ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

A subtending arc on a circle with a radius of 4.5 centimeters has an arc length of 8π. What is the measure of the angle subtende

Romashka-Z-Leto [24]

I hope this helps you

6 0

3 years ago

Elliott is standing at the top of a store escalator that leads to the ground floor below. the angle of depression from the top o

erma4kov [3.2K]

By using what we know about right triangles, we conclude that the height is 11ft.

<h3>How far is the top of the escalator from the ground floor? </h3>

We can think of this as a right triangle.

Where the length of the escalator is the hypotenuse

We know that the angle of depression is 42.51°, then the top angle of our right triangle is:

90° - 42.51° = 47.49°

Now, the height of the top of the escalator would be the adjacent cathetus to said angle, then we can use the relation:

cos(a) = (adjacent cathetus)/(hypotenuse)

Replacing what we know:

cos(47.49°) = height/16ft

cos(47.49°) *16ft = height = 10.8ft

Rounding to the nearest foot, we get:

height = 11ft

If you want to learn more about right triangles:

brainly.com/question/2217700

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4 0

2 years ago

Hello can someone please help?

Ray Of Light [21]

The answer is D
Hope this helps

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3 years ago