All categories

3 years ago

WILL GIVE BRAINLIEST MUST SHOW WORK!

Mathematics

2 answers:

Degger [83]3 years ago

8 0

I THINK IT C

I REALLY TRIED

HOPE I HELPED

Harman [31]3 years ago

5 0

Answer:

Option D is correct.

Step-by-step explanation:

Given: Expenditure in January Month = $ 3365

Expenditure in February Month = $ 3988.50

Expenditure in March Month = $ 4010.02

Expenditure in April Month = $ 4765.87

To find: We need to find average Monthly Expenditures.

$Average=\frac{Sum\:of\:data}{total\:number\:of\:data}$

Average monthly expenditure = $\frac{3356+3988.50+4010.02+4765.87}{4}=\frac{16120.39}{4}=4030.0975$

By rounding fot to nearest hundreths, we get

Average Monthly Expenditure = 4030.10

Therefore, Option D is correct.

You might be interested in

if you rolled a die, what is the probability that you would roll a 4? (simplified fraction as an answer pls)

Ipatiy [6.2K]

The answer would probably be 2/5

4 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago

Which table represents y as a function of x?

tankabanditka [31]

Answer:the top right

Step-by-step explanation:

it only has one y per x

7 0

3 years ago

Read 2 more answers

Please help!!!!!!!!!!!!!!!

konstantin123 [22]

Here's a scenario that you can edit to your own advantage. You own a shirt shop. Each shirt sells for 15 dollars, and to make profit annually you need to sell $62,365 worth of shirts. If you've sold 2,000 shirts so far, how many shirts do you need to meet your quota. How many shirts on average per month would you have to sell?

7 0

3 years ago

If cos -105º = -0.26 and csc -105º = -1.03, then cos 105º = and csc 105º = .

andreyandreev [35.5K]

Answer:

cos (105°) = - 2.6

csc (105°) = 1.03

Step-by-step explanation:

Given that cos (-105°) = - 0.26 {The negative sign is due to the angle - 105° lies in the third quadrant where cos value is negative}

Again, given that csc (- 105°) = - 1.03 {{The negative sign is due to the angle - 105° lies in the third quadrant where csc value is negative}

Now, cos (105°) = - 2.6, because 105° lies in the second quadrant and here cos value is negative.

And csc (105°) = 1.03, because 105° lies in the second quadrant and here csc value is positive. (Answer)

3 0

4 years ago

**WILL GIVE BRAINLIEST** MUST SHOW WORK!

WILL GIVE BRAINLIEST MUST SHOW WORK!