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topjm [15]
3 years ago
9

Describe the Squeeze Theorem as simply as possible.​

Mathematics
1 answer:
Tju [1.3M]3 years ago
5 0

Hey there,

The squeeze theorem shows if f(x)\leq g(x)\leq h(x) for all real numbers (-\infty,~\infty) then f(x) = h(x) but g(x) has to equal that as well.

Let's say we have [ \lim_{n \to \infty} (\frac{cos~x}{x}) ]

~Apply the theorem

Note that [ -1 \leq cos~x \leq  1 ] and [ \lim_{n \to \infty} (-\frac{1}{x})\leq \lim_{n \to \infty} (\frac{cos~x}{x})\leq \lim_{n \to \infty} (\frac{1}{x}) ]

~Apply the infinity property to every side but the middle

\lim_{n \to \infty} (-\frac{1}{x}) = 0

\lim_{n \to \infty} (\frac{1}{x}) = 0

So... \lim_{n \to \infty} (\frac{cos~x}{x})=0

Best of Luck!

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Write the complex number in the form a + bi.
bonufazy [111]

Answer:

\frac{3}{2} + \frac{3\sqrt{3} }{2} i

Step-by-step explanation:

Using the exact value

cos60° = \frac{1}{2} and sin60° = \frac{\sqrt{3} }{2}, then

3(cos60° + isin60°)

= 3( \frac{1}{2} + \frac{\sqrt{3} }{2} i) ← distribute by 3

= \frac{3}{2} + \frac{3\sqrt{3} }{2} i ← in the form a + bi

7 0
4 years ago
Let f and g be linear functions with equations f(x) = m1x + b1 and g(x) = m2x + b2.
MariettaO [177]
F(x) = m1x + b1
g(x) = m2x + b2

f(g(x)) = f(m2x + b2) = m1(m2x + b2) + b1 = m1m2x + b2 + b1

a.) f of g is also a linear function.

b.) slope is m1m2
8 0
4 years ago
Please I need a quick reply​
Rom4ik [11]

Answer:

B

Step-by-step explanation:

It sounds most likely to the right answer

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4 0
3 years ago
Read 2 more answers
Which of the following statements are true regarding functions?<br><br> Check all that apply.
Alex

Answer:

The horizontal line test may be used to determine whether a function is one-to-one.
The vertical line test may be used to determine whether a relation is a function.

Step-by-step explanation:

3 0
2 years ago
The value pie/2 is a solution for the equation 2sin^2x-sinx-1=0<br><br> true or false
Rasek [7]

Answer:

Yes, \frac{\pi }{2} is a solution

Step-by-step explanation:

Recall that sin(\frac{\pi }{2})=1. By replacing \frac{\pi }{2} in the given trigonometric equation, we get 0 (zero) as a result, so this value is a solution:

2*(sin(\frac{\pi }{2} ))^2-sin(\frac{\pi }{2})-1= 2*(1)^2-1-1=2-2=0

6 0
3 years ago
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