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3 years ago

Please answer this question now

Mathematics

1 answer:

Lapatulllka [165]3 years ago

5 0

Answer:

Step-by-step explanation:

$Opposite =u\\Hypotenuse = 12\\Adjacent = 7\\Using\: Pythagoras \: Theorem\\Hypotenuse^2 = Opposite^2+Adjacent^2\\12^2= u^2+7^2\\144=u^2 +49\\144-49=u^2\\95=u^2\\\sqrt{95} =\sqrt{u^2} \\u = 9.74\\\\u = 9.7$

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Find the variable of x. <br> 8x-2=-9+7x

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Answer:

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3 years ago

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26/14 divided by 25/5 =

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Answer:

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Step-by-step explanation:

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3 years ago

A student gets 68 marks n therefore gets 85 percent total marks are?

mihalych1998 [28]

Answer:

There are 80 marks in total.

Step-by-step explanation:

Let the number of total marks be $x$ .

The percentage score of the student can be written as the ratio

$\displaystyle \frac{68}{x} = 85\%$ .

However,

$\displaystyle 85\% = \frac{85}{100}$ .

Equating the two:

$\displaystyle \frac{68}{x} = \frac{85}{100}$ .

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3 years ago

What is 1 1/2 · x = 2 3/4

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3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

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and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for <em>S</em>.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for <em>C</em> :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

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$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

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3 years ago