All categories

1 year ago

Please help me to solve this problem 2q+2q-q

Mathematics

2 answers:

ioda1 year ago

5 0

Answer:

$3q$

Step-by-step explanation:

Hi there,

2q + 2q =4q

From 4q we reduce q which results in :

$3q$

Mark me as brainliest.

gladu [14]1 year ago

4 0

Answer:

Step-by-step explanation:

combine like terms

2q+2q = 4q

4q-q = 3q

You might be interested in

A car dealership pays $8,350 for a car. They mark up the price by 17.4% to get the retail price. What is the retail price of the

tangare [24]

$ 9802.9and it said the answer had to be 20 characters long so i wrote this

3 0

2 years ago

7TH GRADE MATH!!! NEED ASAP

ludmilkaskok [199]

Answer:

l >= 60

Step-by-step explanation:

l greater than or equal to 60

8 0

2 years ago

Read 2 more answers

A car is brought for $28 000.00 and it will depreciate at a rate of $100 per month.

yawa3891 [41]

The ans for a) is 26200$
for b) is 24400$
for c) is 16.3 years

6 0

2 years ago

Yeah so the picture explains it all

Vanyuwa [196]

Answer:

1/3

Step-by-step explanation:

4 0

2 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

2 years ago