All categories

3 years ago

Identificar todas las formas que se representa el intervalo del dominio de la función

Mathematics

1 answer:

leonid [27]3 years ago

7 0

Answer:

III

Step-by-step explanation:

You might be interested in

Mr. Sanchez bought 2 magazine for 9.95 each and 1 book for 14.95 if the sales tax is 6% what is the total cost of Mr. Sanchez pu

Dima020 [189]

Rst you have to find the price of everything together

2 magazines for 9.95 each and a book for 14.95

9.95 + 9.95 + 14.95 = 34.85

Next you convert the percent to a decimal

6% = 0.06

Now you put a one in the ones place

1.06

Multiply this by the cost of everything

34.85 * 1.06 = 36.94

The answer is $36.94

Hope this helps!

Read more on Brainly.com - brainly.com/question/10235412#readmore

5 0

3 years ago

The expression below is scientific notation forwhat number? 4.29*10^-5

notsponge [240]

.00429
I believe this is right! Have a good day!

3 0

3 years ago

Sabrina bought a hoodie at the store when they were having a 25% off sale. If the regular

Rina8888 [55]

Answer:

$12.50

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

which of the following rational functions is graphed below. A.F(x)=1/x(x+2) B. F(x)=2x/x(x-2) C.F(x)=2x/x(x-2) D. F(x)=(x+1)/(x+

nexus9112 [7]

Answer:

you don't have the graph

Step-by-step explanation:

3 0

2 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