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3 years ago

What is 2940 divided by 8

Mathematics

2 answers:

emmainna [20.7K]3 years ago

8 0

367.5, there are 367.5 8's in 2,940.

Hope this helps, if not, comment below please!!!!

In-s [12.5K]3 years ago

8 0

Answer:

365.7

Step by step solving:

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6q + 8 = -9 - 76 a. 12 b. -12 c. 84 d. -84

Alona [7]

Answer: the answer will be -15.5, but this number is not in the choices given, so I guess you made a mistake while writing the equation

Step-by-step explanation:

6q+8=-9-76

6q+8=-85 Subtract 8 from both sides

6q=-93 Divide both sides by 6

q=-15.5

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3 years ago

In one year Megan spent 359.94 on her gym membership at the same monthly rate of 59.99. Write an equation to represent this situ

Thepotemich [5.8K]

Answer:

359.94 = 59.99 * X

Step-by-step explanation:

The situation is determined by the following equation:

T = M * X

Being M the monthly value, that is to say 59.99

T the total value, that is 359.94

And X would be the number of months, and at the same time the variable.

In this case it would be like this:

359.94 = 59.99 * X

If we solve for X, we are left with:

X = 359.94 / 59.99

X = 6

That is to say that Megan went to the gym for 6 months of the year

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4 years ago

The heights of adult men in America are normally distributed, with a mean of 69.1 inches and standard deviation of 2.65 inches.

Allisa [31]

So step by step explanation

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3 years ago

Find the product of (x − 7)2 and explain how it demonstrates the closure property of multiplication.

Ksivusya [100]

Answer:

Step-by-step explanation:

(x-7)^2=(x-7)(x-7)=x^2-7x-7x+49=x^2-14x+49

3 0

3 years ago

Read 2 more answers

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:

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

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

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

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

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

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

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

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

$y(40) = 144 - 33.12693299$

$y(40) = 110.873 \ kg$

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4 years ago