All categories

3 years ago

Solve for f d=16ef^2

Mathematics

1 answer:

vitfil [10]3 years ago

6 0

Here are a bunch of CORRECT answers. Your answer is in the first pic. I got number 3 wrong, but it still showed the correct answer.

You might be interested in

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

Need help with this plzzz

beks73 [17]

Line 1

Natalie was supposed to use PEMDAS (as we all do when doing math), which stands for: Parentheses, Exponents, Mulitplication, Division, Addition, Subtraction. However, Natalie subtracted 1 from 2 before multiplying 6 by 4 and dividing it by 2.

Hope this helps!!

3 0

3 years ago

PLEASE RESPOND QUICKLY IK ON A TIMER

dsp73

Answer: Third one.

Step-by-step explanation:

5 0

3 years ago

The following data represent the pulse rates? (beats per? minute) of nine students enrolled in a statistics course. Treat the ni

aleksandr82 [10.1K]

Answer:

a) $population\hspace{0.1cm} mean = \frac{64 + 77 + 89 + 69 + 89 + 65 + 88 + 69 + 87}{9} = 77.44$

b) $sample\hspace{0.1cm} mean 1 = \frac{65 + 69 + 77}{3} = 70.3$

$sample\hspace{0.1cm} mean 2 = \frac{64 + 69 + 77}{3} = 70$

c) both the means of the two samples underestimate the population mean.

the mean pulse rate of sample 1 underestimates the population mean.

the mean pulse rate of sample 2 underestimates the population mean.

Step-by-step explanation:

i) population size, N = 9

a) $population\hspace{0.1cm} mean = \frac{64 + 77 + 89 + 69 + 89 + 65 + 88 + 69 + 87}{9} = 77.44$

b) $sample\hspace{0.1cm} mean 1 = \frac{65 + 69 + 77}{3} = 70.3$

$sample\hspace{0.1cm} mean 2 = \frac{64 + 69 + 77}{3} = 70$

c) both the means of the two samples underestimate the population mean.

the mean pulse rate of sample 1 underestimates the population mean.

the mean pulse rate of sample 2 underestimates the population mean.

3 0

3 years ago

What is the independent variable of an equation?

wolverine [178]

X is. Y is the dependent variable cause most of the time u plug in x.

6 0

3 years ago