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2 years ago

I get it and then I don’t. I need some refreshing

Mathematics

1 answer:

zavuch27 [327]2 years ago

3 0

Check the picture below.

the length cannot be a negative value, so is not -13.

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Given the function ƒ(x) = x 2 - 4x - 5 Identify the zeros using factorization. Draw a graph of the function. Its vertex is at (2

s344n2d4d5 [400]

The zeroes of the equation are 5, -1

6 0

2 years ago

The function P(x)equals0.35xnegative 76 models the relationship between the number of pretzels x that a certain vendor sells an

r-ruslan [8.4K]

Answer:

$P(X=700) =0.35*700 -76= 169$

Step-by-step explanation:

For this case we have the following profit function:

$P(X) = 0.35 X -76$

And we are interested on find the profit after selling 700 pretzels, so we just need to replace X=700 into the profit function like this:

$P(X=700) =0.35*700 -76= 169$

The profit function is on the figure attached. We see that for 0<X<217 we have negative profits, and for X>217 we will have positive values for the profit. And also we can see that for X=700 we have a profit of approximately 169.

3 0

3 years ago

Someone please help me with this chart, I don’t understand it.

lidiya [134]

Answer:

The second one is (3 x 3 x 3 x 3) x (3 x 3)

Step-by-step explanation:

6 0

3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago

People help please ?????

Mila [183]

Answer:

ok so anything raised to the power 0 is equal to 1 so we'll take 9x^0 =1 and then let's solve

Step-by-step explanation:

3x^2(9x)^0

3x^2(1)

=3x^2

4 0

3 years ago