HELP PLEASE!!! f(0)=4, f(n)=2f (n-1) write the first six terms of the sequence

What is the slope of the line that passes through the points (9, 4)(9,4) and (3, 9)(3,9)? Write your answer in simplest form.

nalin [4]

Answer:

I’m guessing the double points is a typo.

Slope: -5/6

Step-by-step explanation:

Hope this helps <3

Btw love your username <333

6 0

2 years ago

Read 2 more answers

What is the definition of mean???

Scilla [17]

Mean, in terms of math, is the total added values of all the data in a set divided by the number of data in the set. Make sense? If not, here' an example...

Let's say this is my data set:
1, 2, 5, 4, 3, 8, 7, 4, 6,10

To find the mean...
Step 1: Add all of them together.
1+2+5+4+3+8+7+4+6+10 is what? 50. Now that you have this number...
Step 2: Divide by the amount there are. Basically, count up all of the numbers. How many are there? There are 10. Finally...
Step 3: Divide. 50/10 is 5, so the mean of this data set would be 5. Get it? I sure hoped this helped :)

3 0

3 years ago

How to do this question plz

enyata [817]

Answer:

x = 10

Step-by-step explanation:

Use the Pythagorean theorem. The sum of the square of the sides is the square of the hypotenuse.

x² +(√200)² = (√300)²

x² = 300 -200

x = √100 = 10

The length of the unknown side is 10 units.

3 0

3 years ago

José sale de su casa con $50 y gasta 4/5 en el cine y 1/10 en chocolates, ¿qué fracción del

netineya [11]

Answer:

9/10

Step-by-step explanation:

8 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago