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Arada [10]
3 years ago
9

Carlota has 3/4 ton of mulch she is going to divide evenly among 5 flower beds. How much mulch will each flower bed contain? Ple

ase show work
Mathematics
1 answer:
Murljashka [212]3 years ago
5 0
The first thing to do is to turn 5 into a fraction.  
5 = 5/1

Now it is just a simple fractional division question: 
3/4 / 5/1

With fractional division, all you do is switch around the fraction that is on the right and turn the divide sign into a multiplication sign: 
3/4 X 1/5

Now you simply multiply the numerators together, and the denominators together: 
3 X 1 = 3
4 X 5 = 20

3/20. 

Each flower bed will contain 3/20 of a ton of mulch. 

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4535 divided by 83 = R =
Leto [7]

Answer:

54 R 53

54 53/83

Step-by-step explanation:

4535 divided by 83 equals

54 with a remainder of 53

8 0
3 years ago
Hey guys stuck on three questions, can you help out a bit? Thanks!
alexandr1967 [171]

Answer:

D

Step-by-step explanation:

Because the numbers do not comply with the like 11+6+18+21= no answer and multiplying them all didn't matter either so its none of the above

7 0
3 years ago
Pam and her brother both open savings accounts. Each begin with a balance of zero dollars. For every two dollars that Pam save i
dezoksy [38]

Answer:

Ratio [Pam to her brother] = 2:5

Step-by-step explanation:

Given:

Pam saves every two dollar

Her brother saves fives dollar (For every two dollar)

Find:

Ratio [Pam saving and his brothers saving]

Computation:

It is given that if Pam saves $2 her brother saves $5

So,

Ratio [Pam to her brother] = 2/5

Ratio [Pam to her brother] = 2:5

6 0
4 years ago
Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut
frozen [14]

Answer:

a)\sqrt{144-288t+208t^2} b.) -12knots, 8 knots c) No e)4\sqrt{13}

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved 12\dot t (n.m) and ship B has moved 8\dot t (n.m). We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}

b)

We want to find \frac{ds}{dt} for t=0 and t=1

\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0

Since function f(x)=199-288x+208x^2 is quadratic, concave up and has no real roots, we know that 199-288x+208x^2>0 for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

                                                \lim_{t \to \infty} \frac{ds}{dt}

So, lets check this limit:

\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}

Notice that:

4\sqrt{13}=\sqrt{12^2+5^2}=√(speed of ship A² + speed of ship B²)

5 0
4 years ago
Solve using the histograms. How many watermelons weigh more than any of the cantaloupes? A. 10 B. 14 C. 17 D. 21
AleksAgata [21]
The answer is B. 14.
6 0
4 years ago
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