I need help with this problem

(3.2 + 6.8)x - (36 divided by 4)x PLEASE HELPPP

Mrac [35]

Answer:

i dont no but what i got was 10x+144d^3 but do u want to sovle for x i got u X

Step-by-step explanation:

6 0

2 years ago

Diego has two jobs. He works 3 hours at the library each day, making $8 per hour. In addition, he works at a coffee shop, making

Zepler [3.9K]

Answer:

D. l = 3 || 4 < c < y || 8l + 7.5c ≥ 70

Step-by-step explanation:

<h3>Work at library</h3>

Time = 3 hours ⇒ l = 3
Payment = $8 per hour
Total payment = 8l

<h3>Work at coffee shop</h3>

Time = between 4 and 7 hours ⇒ 4 < c < 7
Payment = $7.5 per hour
Total payment = 7.5c

Target income = at least $70 daily

<u>So the system as per above parameters:</u>

l = 3
4 < c < y
8l + 7.5c ≥ 70

Correct option is D

3 0

2 years ago

An individual head of a sprinkler system covers a circular area of grass with a radius of 25 feet. the yard has 3 sprinkler head

frutty [35]

Answer:

Option D. 5890.5 feet²

Step-by-step explanation:

An individual head of a sprinkler system covers a circular are of grass with a radius of 25 feet.

Yard has three sprinkler heads that each cover circular area with no overlap.

We have to calculate the approximate area watered by these 3 sprinkler heads.

Area covered by one sprinkler = π×r² = π×(25)² = 625π feet²

Therefore area covered by three sprinkler heads = 3 × 625π = 5892.85 feet²

which matches the approximate value of Option D. 5890.5 feet²

3 0

2 years ago

Read 2 more answers

Becky is rolling a number cube with the number 1 through 6 what is the probability she will roll a 4 or 5

ryzh [129]

1/6 since there are 6 sides, therefore, there is a 1/6 chance that any number will be rolled

8 0

2 years ago

Read 2 more answers

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