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nadya68 [22]
2 years ago
5

What is the length of the line segment with endpoints (−3,8) and (7, 8) ?

Mathematics
1 answer:
Savatey [412]2 years ago
5 0
Do you have a picture or is that it?
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dean counted 10 red cards, 10 black catds, and 20 blue cards in a deck of cards. what is the ratio of red cards to all cards?
miv72 [106K]

Answer:

10:40

Step-by-step explanation:

10+10+20=40

10 red cards

4 0
2 years ago
HELP I NEEF UT WITHIN 3mins! PLS HELP GOD BLESS U
Feliz [49]

Answer:

8.0%

Step-by-step explanation:

<u>.8</u> * 5,000 = 400

4 0
2 years ago
Given the following table, find the rate of change between f(-1) and f(2)
stealth61 [152]
I think it’s 7/12 I think it would 7/12
3 0
3 years ago
A large concrete block is used for the cornerstone
marin [14]

Answer:

The volume of the concrete needed to make the block is 5054 ft³

Step-by-step explanation:

The question relate to the calculation of the volume of a shape

The dimensions of the block are;

The depth of the block, d = 19 ft.

The width of the block, w = 19 ft.

The length of the block, l = 14 ft.

The volume of a cube, V = Length, l × Width, w × Depth, d

Therefore, the volume, 'V' of a cubic block is given as follows;

V = l × w × d = 14 ft. × 19 ft. × 19 ft. = 5054 ft.³

The volume of the concrete needed to make the block = The volume of the block, V = 5054 ft³.

8 0
2 years ago
A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min
Dmitry [639]

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

6 0
3 years ago
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