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3 years ago

Please please please please please help me

Mathematics

1 answer:

Lelu [443]3 years ago

6 0

Answer:

nhhhhhhhhhhhhfvghgvbjh

Step-by-step explanation:

nvnbhnbjhnbjnbjjmjm

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Mr. Anderson tosses a coin 15 times and it landed on tails 9 times. What is the probability of landing on tails on his 16th toss

Anestetic [448]

It is a 50 percent chance. This is because there is only two possibilities that can occur. It does not matter how many times the coin has been flipped.

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3 years ago

Find the midpoint of the line segment between the points (11, -5) and (-3, -7)

alexdok [17]

Answer:

(4, -6)

Step-by-step explanation:

(x1+x2)/2 + (y1+y2)/2 = midpoint

(11+-3)/2 = 8/2 = 4. x = 4

(-5 + -7)/2 = -12/2. y = -6

4, -6 is the midpoint

give brainliest please! hope this helps :)

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1 year ago

If a^2 + b^2=c^2 then the triangle is a right triangle.

NARA [144]

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Step-by-step explanation:

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3 years ago

Given a mean of 571 and a standard deviation of 236.5, what is the z-score of the value 724 rounded to the nearest tenth? a.0.9

tangare [24]

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3 years ago

Read 2 more answers

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