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2 years ago

Please please please answer this I have the answer I just need the work for it!!!!!

Mathematics

1 answer:

MrRa [10]2 years ago

3 0

Answer:

x=80

Step-by-step explanation:

First of all, you must find the length of the side to add with 36 to equal 100. This would give you 64. So once you do that, you can do pythagorean theorem to find the length of x. This would give you 64^2 + 48^2 = x^2. Once you have this equation, you square your terms to get; 4,096 + 2,304 = x^2. After you add them you will now have 6,400 = x^2. Once you have this, you will need to do the square root of both sides of the equation; sqrt(6400) = x. This gives you x = 80.

Hope this helps!

Brainliest?

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4(9a2−6a+1)
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1 year ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

10 3/9- 10 2/5 Im to lazy to do it please help me out.

coldgirl [10]

Answer:

$\boxed{ \ - \frac{3}{45} }$

Step-by-step explanation:

10 3/9 - 10 2/5

we need to convert mixed fractions to improper fractions

$10 \frac{3}{9} = \frac{10 \times 9 + 3}{9} = \frac{93}{9}$

$10 \frac{2}{5} = \frac{10 \times 5 + 2}{5} = \frac{52}{5}$

Now, we have 93/9 - 52/5

We need to equate the denominators, before subtracting them.

$\frac{93}{9} - \frac{52}{5} = \frac{93 \times 5}{45} - \frac{52 \times 9}{45}$

$\frac{465}{45} - \frac{468}{45}$

$\frac{ - 3}{45}$

$\boxed{ - \frac{3}{45} }$

The result of 10 3/9- 10 2/5 is - 3/45

8 0

1 year ago

State which of the following sets of ordered pairs represent a function.

zalisa [80]

Answer:

Option (D) is correct.

Step-by-step explanation:

A function can be termed as a process from a set of x values to a set of possible y-values where each x-value relates to exactly one y-value.

In other words, a relation can only be a function there is no repetition of x-values. It means a relation can not have duplicated inputs.

Given the sets

Set A: (5,2), (4, 3), (3, 4), (2,5)

Set B: (-1,-6), (0, 2), (1, 2), (3, 6)

Set C: (2, 1), (4, 2), (2, 3), (8,4)

It is clear that sets A and B represent the function because, in these sets, each x-value relates to exactly one y-value. In other words, there are no duplicated inputs.

But, set C has duplicated inputs i.e. x = 2 is repeated twice, and a function can not have a repeated input.

Therefore, we conclude that set A and set B represent a function.

Thus, option (D) is correct.

7 0

2 years ago