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3 years ago

Can someone help me?

Mathematics

1 answer:

Katena32 [7]3 years ago

4 0

Answer:

y=2x+0

Step-by-step explanation:

you make $2 per day. that is your rate of change or m.

the intercept is 0 because you get no money if you don't work.

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Shoemakers of America forecasts the following demand for the next six months: 5000 pairs in month 1; 6000 pairs in month 2; 7000

Andrew [12]

Answer:

5,000+6,000+7,000+,9,000+6,000+5,000= 38,000

Step-by-step explanation:

8 0

3 years ago

How much is 1 foot line of quarters worth

Alborosie

a quarter is 1.75 mm thick

1 mm = 0.0393701 inches

0.0393701/12 = 25.4 mm per inch

25.4/1.75 = 14.5 quarters per inch

14.5 x 12 = 174 quarters per foot

174 x 0.25 = 43.50

$43.50 for a foot of quarters

5 0

4 years ago

Please could I have some help? Seven squared equals seven times .........

viva [34]

Answer:

$\large \boxed{\mathrm{seven }}$

Step-by-step explanation:

$7^2$

A number squared also means that the number is multiplied by itself twice.

$7^2 =7 \times 7$

4 0

3 years ago

Read 2 more answers

11 ft<br>15 ft<br>13 ft what is the volumn

MakcuM [25]

Answer:

2145 ft^3

Step-by-step explanation:

7 0

3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago