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3 years ago

What is the product of 7x-4 and 8x+5

Mathematics

1 answer:

Romashka [77]3 years ago

4 0

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ (7x - 4)(8x + 5) ==> 56x^2 + 35x - 32x - 20

This can be simplified to 56x^2 + 3x - 20

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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What is they answer on the sheet of paper for my homework

geniusboy [140]

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3 years ago

A digital camera costs $185 and the sales tax rate is 5.7%. What is the total cost of the camera after tax

Dovator [93]

♥ To solve find how much this will cost AFTER the price increases.
♥ Solve:
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3 years ago

Lim (n/3n-1)^(n-1) n → ∞

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, e :

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

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

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

From the relation 9m = n - 9, we see that m also approaches infinity as n 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:

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

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3 years ago

Review the steps of the proof of the identity

koban [17]

Answer:

step 2

and then also in step 3 compensating the error in step 2

Step-by-step explanation:

I think I just answered this for another post.

sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

so, step 1 is correct :

sin(A - 3pi/2) = sin(A)cos(3pi/2) - cos(A)sin(3pi/2)

but step 2 suddenly and incorrectly switched that central "-" to a "+".

yes, sin(3pi/2) = -1, but that is still an explicit factor in step 2. so it was not used to flip the central operation from subtraction to addition, and therefore this change was a mistake.

then, in step 3, another error was made by just ignoring the "-" sign of "-1" and still keeping the central "+" operation. this error compensated for the error in step 2 bringing us back by pure chance to the right result.

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2 years ago

A coin is loaded so that the probability of heads is 0.55 and the probability of tails is 0.45. suppose the coin is tossed twice

Vladimir79 [104]

The probability is 0.235

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3 years ago