there is an average of 16 students per teacher at the middle school if 192 students go to the school how many teachers are there

Which algebraic expression is a polynomial with a degree of 5

anygoal [31]

Answer: For example, the polynomial which can also be expressed as has three terms. The first term has a degree of 5 (the sum of the powers 2 and 3), the second term has a degree of 1, and the last term has a degree of 0. Therefore, the polynomial has a degree of 5, which is the highest degree of any term.

Step-by-step explanation: (If this helps)

8 0

3 years ago

Read 2 more answers

I need help with this problem, could you help me?

taurus [48]

Fog(x)= f(g(x))

g(x) = 2x^2 - 7x - 4
f(x) = x+4

fog(x) = (2x^2 - 7x - 4) + 4
fog(x) = 2x^2 - 7x

hope it helps

3 0

4 years ago

1. Suppose y varies directly with x, and y = 14 when x = -4. What is the value of y when x = -6

In-s [12.5K]

direct variation

y = kx

substitute in and solve for k

14 = k(-4)

14/-4 =k

k = -7/2

y = -7/2 x

y = -7/2 * -6

y = 42/2 = 21

Choice D

8 0

3 years ago

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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

4 years ago

What is the measure of the complement of 27 degrees?

serg [7]

The answer is: " 63° " .
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Note: "90 <span>− 27 = 63 " .
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6 0

3 years ago