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3 years ago

Please help me this I’m running out of time

Mathematics

1 answer:

Ivan3 years ago

6 0

Answer:

I chose 7 as the base and 2 as the Exponent

Step-by-step explanation:

7^2 . 7 is the base and 2 is the Exponent.

Standard Form: 7^2

Expanded Form : 7 • 7 = 49

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A chess club with 40 members is electing a new president. Hong received 36

Blababa [14]

36/40 = 9/10
and convert into percentage would be (9/10 )*100 which would be 90

6 0

3 years ago

Read 2 more answers

You roll a 6 sided dice what is the probability of rolling a number greater than two and then rolling a number less than three

Alexandra [31]

First answer is 4 chances and second answer is 2 chances

5 0

4 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

Rational number with a denominator of 4 that fall between -1 and -2

asambeis [7]

Answer:

-5/4

Step-by-step explanation:

You can use this it is -1.25 or -5/4, and lies between -1 and -2, rational numbers are numbers that can be expressed in the form x/y of two integers.

6 0

3 years ago

All fractions equivalent to 27/81

Strike441 [17]

There is more than a trillion of equivalent fractions to 27/81, you just need to divide of multiply that number to any other numbers, but not 1 and 0, it can be negative number also
Ex: 1/3, 3/9, 54/12, 81/243, etc...

3 0

4 years ago