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3 years ago

Odette solved this problem:

Mathematics

1 answer:

Tju [1.3M]3 years ago

3 0

Answer:

In Odette's solution, 7 is the number of

nickels and 8 is the number of dimes

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You have five quarts of paint. It takes 5/8 quart to paint each chair you are asked to paint. How many chairs can you paint? sho

ANTONII [103]

Answer:

8 chairs

Step-by-step explanation:

5/(5/8)=8

7 0

3 years ago

Read 2 more answers

Pleaseeee help!!!! I will mark you as brainlinest for correct answer!!!!

Yuri [45]

I think it would be two units up and 1 unit to the left.

If you have any more questions, let me know. <3

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3 years ago

A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.8kg/L

Scorpion4ik [409]

Answer:

1. $\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}$

2. $y(40) = 110.873 \ kg$

Step-by-step explanation:

Given that:

A very large tank initially contains 100 L of pure water.

Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.

. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L

t = (50 + t ) L

Since it is given that we should consider y(t) to be the amount of salt (in kilograms) in the tank after t minutes.

Then , the differential equation that y satisfies can be computed as follows:

$\dfrac{dy}{dt}=rate_{in} - rate_{out}$

$\dfrac{dy}{dt}=(0.8)(5) -\dfrac{y(t)}{100+2t} \times3$

$\dfrac{dy}{dt}=(0.8)(5) -\dfrac{3y}{100+2t}$

$\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}$

How much salt is in the tank after 40 minutes?

So,

suppose : $e^{\int \dfrac{3}{100+2t} \ dt} = (t+50)^{3/2}$

Then ,

$( t + 50)^{3/2} y' + \dfrac{3}{2}(t+50)^{1/2} y = 4(t+50)^{3/2}$

$( t + 50)^{1.5} y' + \dfrac{3}{2}(t+50)^{0.5} y = 4(t+50)^{3/2}$

$[y\ (t + 50)^{1.5}]' = 4(t+ 50)^{1.5}$

Taking the integral on both sides; we have:

$[y(t + 50)^{1.5}] = 1.6 (t + 50)^{2.5} + C$

$y = 1.6 (t+50)+C(t+50)^{-1.5}$

$y(0) = 0 = 1.6(0+50) + C ( 0 + 50)^{-1.5}$

$0 = 1.6(50) + C ( 50)^{-1.5}$

$C= -1.6(50)^{2.5}$

$y(40) = 1.6 (40 + 50)^1 - 1.6 (50)^{2.5}(50+40)^{-1.5}$

$y(40) = 144 - 1.6 \times 17677.66953 (90)^{-1.5}$

$y(40) = 144 - 1.6 \times 17677.66953 \times 0.001171213948$

$y(40) = 144 - 33.12693299$

$y(40) = 110.873 \ kg$

6 0

4 years ago

If AD=8 and BD=4, what is the length of DC⎯⎯⎯⎯⎯⎯?

balu736 [363]

Answer:

the answer should be 16!

4 0

3 years ago

d. Is the statement "Two fundamental questions about a linear system involve existence and uniqueness" true or false? Explain

Roman55 [17]

Answer:

True

Step-by-step explanation:

Because two fundamental questions address whether the solution exists and whether there is only one solution.

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4 years ago