Which is equal to 7√3 ? 7^3 21 (13)^7 7 1/3

The sum of 5 consecutive integers is 120. What is the third number in this sequence?

aliina [53]

Ok so

x - the smallest
The other 4 are : x+1, x+2, x+3, x+4
We know that sum is 120
So
x+x+1+x+2+x+3+x+4=120
Combine like terms
5x+10=120
Subtract 10 on both sides
5x=120-10
5x=110
Divide by 5 on both sides
X=22
The third number is x+2 = 22+2=24

3 0

3 years ago

I don’t get this...!!!

notka56 [123]

Ray CB, because it will include point A in the ray
Rays go on forever in one direction, so Ray CB and Ray CA are the same because they're going in the same direction

3 0

4 years ago

An operational definition is used to ____ a hypothetical construct.

Alona [7]

An operational definition can both define and measure the hypothetical construct. The hypothetical construct is also called psychological construct is an explanatory variable which cannot be directly observed. It is also a tool to understand someone’s behavior. For example measuring you friend happiness.

8 0

4 years ago

Read 2 more answers

First person with a answer is marked as the Brainliest Answer

kramer

Answer:

h(x) = x/5

Step-by-step explanation:

We want to know the value of y to satisfy the given equation. So:

$6x+y=4x+11y\\\\y-11y=4x-6x\\\\-10y=-2x\\\\10y=2x\\\\y=\dfrac{2x}{10}\\\\\boxed{y=\dfrac{x}{5}}$

Hence, since h(x) returns the value of y we found before:

$\boxed{h(x) = \dfrac{x}{5}}$

3 0

3 years ago

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