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IceJOKER [234]
3 years ago
9

sex(-x)-sin(-x)tan(-x)=cosx Note that each Statement must be based on a Rule chosen from the Rule menu. To see a detailed descri

ption of a Rule, select the More Information Button to the right of the Rule.
Mathematics
1 answer:
julsineya [31]3 years ago
4 0

Answer:

Step-by-step explanation:

sec (-x)-sin (-x) tan (-x)=cos x

sec(x)-(-sin x)(-tan x)=cos x

1/cos x-sin x(sin x/cos x)=cos x

multiply by cos x

1-sin ²x=cos²x

or 1=sin²x+cos²x

which is true.

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Plz help and explain thanks!!!
Mars2501 [29]
The answer to the question

3 0
2 years ago
You decide to put $100 in a savings account to save for a $3,000 down payment on a new car. If the account has an interest rate
saul85 [17]
By using the formula: 

<span>FV = PV*(1+r/n)^(n*t)

</span><span>FV = 3000
PV = 100
 r = 0.02
 n = 12
t = ?</span>

Substituting the values of the variables, I got <span>a. 170.202 years. Hope that really helps. </span>
8 0
3 years ago
Determine if x=-16 is a zero &amp; find the quotiant and the remainder​
DerKrebs [107]

Answer:

○ A. No, x = -16 is not a zero of the polynomial.

The quotient is x² - 2x + 83, and the remainder is -1274.

Step-by-step explanation:

This is not a zero. When set to equal zero, you get these roots: -2, -3, -9. Now, about the the Remainder Theorem, I have not been taught this, but just by looking at it, I come across this as the remainder.

I am joyous to assist you anytime.

* I cross my fingers for it to be correct!

3 0
3 years ago
Tiva earns $48 for 6 hours of babysitting. Complete each statement if Tiva keeps earning her babysitting money at this rate. CLE
lawyer [7]
8.5
= (48 \div 6) \times 8.5
8 \times 8.5
= 68

$32

Perhour =8
32/8 =4 hours
3 0
3 years ago
Read 2 more answers
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

7 0
3 years ago
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