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2 years ago

What is sandwich theorem?

2 answers:

7 0

In calculus, the squeeze theorem, also known as the pinching theorem, the sandwich theorem, the sandwich rule, the police theorem, the between theorem and sometimes the squeeze lemma, is a theorem regarding the limit of a function. In Italy, the theorem is also known as theorem of carabinieri.

Novay_Z [31]2 years ago

6 0

In calculus, the squeeze theorem, also known as the pinching theorem, the sandwich theorem, the sandwich rule, the police theorem, the between theorem and sometimes the squeeze lemma, is a theorem regarding the limit of a function. In Italy, the theorem is also known as theorem of carabinieri.

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Please help with 24-26

Inga [223]

The power 0 only affects the x. x^0 = 1

-2 * 1 = -2

Here everything inside the brackets is affected by the 0.

(11abx)^0 = 1 There is nothing left over.

(xy)^0 = 1

(pq)^0 = 1

1 - 1 = 0

3 0

2 years ago

Read 2 more answers

Find the values of x and y

qaws [65]

Answer:

y=30

x=60

Step-by-step explanation:

The triangle has all sides equal therefore all the angles are equal.

For all the angles to be equal and add up to 180 (sum of angles in a triangle is 180), each angle has to be 60.

Therefore y=90-60

=30

x=180-(30+90)

=60

8 0

2 years ago

Josh has $25 and he wants to buy some movies. He picks one movie that costs $9.99 and two that cost $5.99 each.

vova2212 [387]

He should add exact since money is involved
he could also estimate since they are so close together, but he would be off only 2 cents

A

3 0

3 years ago

Read 2 more answers

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

2 years ago

Solve the following equations for 0 ≤ theta ≤ 360º cosec(2 — 45°) = 2

lions [1.4K]

Step-by-step explanation:

here ,

$cosec \alpha = \frac{1}{ \sin( \alpha ) }$

now,

cosec(2-45°)=2

or,

1/sin(2-45°) =2

or,

1/sin2cos45°-cos2sin45=2

or,

$\frac{1}{ \sin(2) \frac{1}{2} - \cos(2) \frac{1}{2} } = 2$

or,

sorry I have that much qualifications to work on this question and hoping this much will make bit easy for you to solve it further

4 0

1 year ago