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3 years ago

Can someone helppp plss show work

Mathematics

2 answers:

Vladimir79 [104]3 years ago

3 0

Answer:

m = 3/5

b = -5

Step-by-step explanation:

Equation for y-intercept form is y=mx + b

No math is needed, you have the answer already

faust18 [17]3 years ago

3 0

put the number in y=mx+b form

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3 years ago

One pound of chicken cost $3.00. Jim buys 8 2/3 pounds of chicken. How much does Jim pay for the chicken?

DaniilM [7]

It would cost 26 dollars because 3 multiplied by 8 is 24 plus 2/3 of 3 is 26 and that is the result of 2/3 of 3 being 2 dollars

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4 years ago

John has × marbles and max has twice as many.max give gives john 5 of his marbles does max now have

rjkz [21]

Answer: 2x -5

Step-by-step explanation:

Hi, to answer this question we have to write equations with the information given:

John has × marbles

John=x

Max has twice (multiplied by 2) as many.max give gives john 5 of his marbles.

We have to multiply by 2 the number of marbles that John has (x), and subtract 5.

Max = 2x -5

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

A mathematical phrase represented by numbers, variables, and operations is a(n)

Andre45 [30]

Answer:

Expression

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Expressions use numbers, variables, and operations to form a mathematical calculation.

Equations are similar except they use equal signs, so that makes them different than expressions.

If this answer is correct, please make me Brainliest!

7 0

4 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:

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

7 0

3 years ago