All categories

3 years ago

The sum of twice a number and 4 is equal to the sum of the number 6

Mathematics

2 answers:

Bad White [126]3 years ago

7 0

Answer:

n * 2 + 4 = n + 6

Step-by-step explanation:

Sum of twice a number and 4 is n * 2 + 4

Equal to the sum of number 6 is = n + 6

Vlad1618 [11]3 years ago

6 0

What kind of evil school are you going to?

If I am getting this right, the answer is 1. I think it’s asking what number plus itself plus four = 6. It says,”equal to the sum of the number 6” which I guess is just 6.

So I believe the answer is 1

You might be interested in

At a sandwich shop, they charge $2.75 for the basic sandwich and then $0.45 per ounce for the toppings. If

riadik2000 [5.3K]

She has 5 ounces of toppings.
The reasoning is you first do 5-2.75 because you must subtract the basic sandwich from the total cost. Then, you must divided 2.25/0.45 because that is the cost of the remaining toppings.

7 0

4 years ago

Read 2 more answers

Which ordered pairs make both inequalities true? Check all that apply.

Thepotemich [5.8K]

All others did not satisfy the inequality simultaneously except #4 and #6
For #4:
2 < 5(1) + 2 i.e. 2 < 7 (true)
2 >= 1/2(1) + 1 i.e. 2 >= 3/2 (true)
For #6
2 < 5(2) + 2 i.e. 2 < 12 (true)
2 >= 1/2(2) + 1 i.e. 2 >= 2

7 0

3 years ago

Read 2 more answers

#6<br> What do I do with the () in the question

GuDViN [60]

That just means that you are dividing by -3 so when you do 45 divided by -3 what do you get?

7 0

3 years ago

Carlos has 10 more dollars and Jeremy Jeremy has $5 more than the show all together they have $80 what part of 80 does Michelle

blsea [12.9K]

She would have 60 dollars.

5 0

3 years ago

2) Use spherical coordinates to find the integral of f(x,y,z) = z over x^2 + y^2 + z^2 < 4 and x,y,z < 0

777dan777 [17]

1. Denote by $\mathcal D$ the region bounded by the cylinder $x^2+y^2=9$ and the planes $z=0$ and $z=6$ . Converting to cylindrical coordinates involves setting

$\begin{cases}x=r\cos u\\y=r\sin u\\z=v\end{cases}$

and $\mathcal D$ is obtained by varying the parameters over $0\le r\le3$ , $0\le u\le2\pi$ , and $0\le v\le6$ . The volume element is

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}xz\,\mathrm dV=\int_{v=0}^{v=6}\int_{u=0}^{u=2\pi}\int_{r=0}^{r=3}r^2v\cos u\,\mathrm dr\,\mathrm du\,\mathrm dv=0$

2. Now let $\mathcal D$ denote the region bounded by the sphere $x^2+y^2+z^2=4$ and the coordinate planes in the octant in which each coordinate of the point $(x,y,z)$ is negative.

Converting to spherical coordinates, we set

$\begin{cases}x=\rho\cos u\sin v\\y=\rho\sin u\sin v\\z=\rho\cos v\end{cases}$

where we obtain $\mathcal D$ by varying over $0\le\rho\le2$ , $\pi\le u\le\dfrac{3\pi}2$ , and $\dfrac\pi2\le v\le\pi$ . The volume element is now

$\mathrm dV=\rho^2\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}z\,\mathrm dV=\int_{v=\pi/2}^{v=\pi}\int_{u=\pi}^{u=3\pi/2}\int_{\rho=0}^{\rho=2}\rho^3\cos v\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv=-\pi$

8 0

3 years ago