Ask question

Ask question

All categories

3 years ago

14

A group of friends were working on a student film. They spent $712 on costumes, which was 89% of their total budget. What was th

e total budget for their student film?

1 answer:

Anna11 [10]3 years ago

8 0

Answer:

800

Step-by-step explanation:

you divide 712 by .89

You might be interested in

Please help me. I need help ASAP.

lara [203]

Answer:

a) 72

b) $259.2

Step-by-step explanation:

A- The bus travels 40 miles on 8 gallons of gasoline. The bus is traveling 360 miles in total.

360/40= 9

9*8= 72

B- $3.60 per gallon and 72 gallons in total.

3.60*72=259.2

$259.2

5 0

3 years ago

Two lines intersect in more than one point.

statuscvo [17]

That would be false because parallel lines don't intersect. But i can if one is straight and the other is Curved

7 0

3 years ago

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

Translate the phrase to an algebraic expression.

mario62 [17]

Answer:

A

Step-by-step explanation:

If it’s 80 less than, your subtracting and then 4 times a number is 4x so your answer is 4x-80

8 0

3 years ago

Read 2 more answers

Neeeedddd Helpppp Plzzzzz

Nataly_w [17]

Answer:

Linear Angles

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

4 0

3 years ago

Read 2 more answers