All categories

3 years ago

Solve for q: 6n-5p/11t=c

1 answer:

6 0

Your question can be interpreted in two ways, I will solve both you then take the one you mean. Also there is no q in your equation, but I suppose you mean p.
$\frac{6n-5p}{11t} =c \\ 6n-5p=11ct \\ 6n-11ct=5p \\ p= \frac{6n-11ct}{5}$
Alternatively,
$6n-
 \frac{5p}{11t} =c \\ 66nt-5p=11ct \\ 66nt-11ct=5p \\ p= 
\frac{66nt-11ct}{5}$

You might be interested in

Bobby is making a cake. The recipe calls for 1( 3)/(4) cups of flour for every( 1)/(8) cups of sugar. How many cups of flour is

PilotLPTM [1.2K]

Answer:

5 cups

Step-by-step explanation:

To get 1 cup of sugar, u must add 1/8 by 8. now you have one cup of sugar. But to find out how much flour u need, u also need to multiply 1 3/4 by 8, which gives u 5 cups

5 0

3 years ago

Read 2 more answers

Please help

OlgaM077 [116]

First, we're going to see the candies per minute that machine C packs.
150 / 2 = 75

Now that we know that machine C packs 75 candies per minute, we're going to multiply the candies machine C makes by 11 minutes.
75 x 11 = 825

We're now gonna do the same with machine D.
130 x 11 = 1430

Then we're going to find the difference between machine C and machine D, we do this because the question basically asks how much more candies can machine D pack than machine C.
1430 - 825 = 605 candies.

This is how we find our final answer.

Our answer would be D) 605.

Hope this helps!~

4 0

3 years ago

Mr. Lemus washed

Georgia [21]

Answer:

his son

Step-by-step explanation:

5 0

3 years ago

The Graph of a function g is shown below. Find the following: g(10)

Yanka [14]

Answer: the graph is not shown, but I'll try to explain how hypothetically, you would be able to graph g(10). If you are given something like an image of the graph, find slope and y-intercept first. The y-intercept is the point (0, #), and is on the y-axis. After finding what g(x) = mx + b <- slope intercept form is, substitute 10 for x, then solve.

Step-by-step explanation:

3 0

3 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago