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3 years ago

Please sum1 help? I don’t understand this it would mean the wrld thx

Mathematics

2 answers:

GrogVix [38]3 years ago

8 0

Answer:

Step-by-step explanation:

lord [1]3 years ago

7 0

Answer:

20 pages per hour

Step-by-step explanation:

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In hopes of encouraging healthier snacks at school, Wanda brought in a tray of carrot sticks and apple slices to share. The tray

lana66690 [7]

Answer:

Wanda brought 9 carrot sticks.

Step-by-step explanation:

Convert 30% to a decimal by removing the percent sign and dividing by 100: 30 ÷ 100 = 0.30
Substitute 0.30 for 30% in the equation: 30% x 30 = Y becomes 0.30 * 30 = Y
Do the math: 0.30 x 30 = 9
Y = 9
So 30% of 30 is 9
Double check your answer with the original question: How many carrot sticks did Wanda bring? Multiply 0.30 x 30 = 9

4 0

2 years ago

Help please help me I need this

mixer [17]

Answer:

$\frac{5+-\sqrt{53} }{2}$

Step-by-step explanation:

in the equation given these are the values:

a=1 b= -5 c= -7

if we plug this in the quadratic formula $\frac{-(-5)+-\sqrt{(-5)^2-4(1)(-7)} }{2(1)}$

this is simplified to $\frac{5+-\sqrt{53} }{2}$

hope this helped

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3 years ago

What is the least common multiple (LCM) of 4 and 12?

AleksandrR [38]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Which of the following is a zero of the function f(x) = x3 - x2 - 21x + 45.

marin [14]

The zeros are 3 and -5. Hope this helps!

5 0

3 years ago

A tank with capactity 500 gal originally contains 200 gal water with 100 lbs of salt mixed into it. Water containing 1 lb of sal

prisoha [69]

Answer:

The amount of salt in the tank at any moment $t$ is

$f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

The concentration of salt in the tank when it is at the point of overflowing is $0.968$ .

The theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

Step-by-step explanation:

Let $f(t)$ be the amount of salt in the tank at any time $t.$

Then, its time rate of change, $f'(t)$ , by (balance law).

Since three gallons of salt water runs in the tank per minute, containing $1$ lb of salt, the salt rate is

$3.1=3$

The amount of water in the tank at any time $t$ is.

$200+(3-2)t=200+t,$

Now, the outflow is $2$ gal of the solution in a minute. That is $\frac 2{200+t}$ of the total solution content in the tank, hence $\frac 2{200+t}$ of the salt salt content $f(t)$ , that is $\frac{2f(t)}{200+t}$ .

Initially, the tank contains $100$ lb of salt,

Therefore we obtain the initial condition $f(0)=100$

Thus, the model is

$f'(t)=3-\frac{2f(t)}{200+t}, f(0)=100$

$\Rightarrow f'(t)+\frac{2}{200+t}f(t)=3, f(0)=100$

$p(t)=\frac{2}{200+t} \;\;\text{and} \;\;q(t)=3$ Linear ODE.

so, an integrating factor is

$e^{\int p dt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2$

and the general solution is

$f(t)(200+t)^2=\int q(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac 1{(200+t)^2}\int 3(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac c{(200+t)^2}+200+t$

Now using the initial condition and find the value of $c$ .

$100=f(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}$

$\Rightarrow c=-4000000=-4\times 10^6$

$\Rightarrow f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

is the amount of salt in the tank at any moment $t.$

Initially, the tank contains $200$ gal of water and the capacity of the tank is $500$ gal. This means that there is enough place for

$500-200=300$ gal

of water in the tank at the beginning. As concluded previously, we have one new gal in the tank at every minute. hence the tank will be full in $30$ min.

Therefore, we need to calculate $f(300)$ to find the amount of salt any time prior to the moment when the solution begins to overflow.

$f(300)=-\frac{4\times 10^6}{(200+300)^2}+200+300=-16+500=484$

To find the concentration of salt at that moment, divide the amount of salt with the amount of water in the tank at that moment, which is $500$ L.

$\text{concentration at t}=300=\frac{484}{500}=0.968$

If the tank had an infinite capacity, then the concentration would be

$\lim\limits_{t \to \infty} \frac{f(t)}{200+t}= \lim\limits_{t \to \infty}\left(\frac{\frac{3\cdot 10^6}{(200+t)^2}+(200+t)}{200+t}\right)$

$= \lim\limits_{t \to \infty} \left(\frac{4\cdot 10^6}{(200+t)^3}+1\right)$

$=1$

Hence, the theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

3 0

4 years ago