All categories

3 years ago

When 2 times a number is subtracted from 7 times the number, the result is 55. what is the number?

1 answer:

8 0

Let n represent the number. You require
7n - 2n = 55
5n = 55 . . . . . . collect terms
n = 11 . . . . . . . divide by 5

The number is 11.

You might be interested in

An 8th Grader is paid for the chores they do around the house. (Although they shouldn't be paid... They don't pay rent and eat f

Solnce55 [7]

Answer

100 to 300

200 to 300

320 to 400

250 to 300

I hope this helps

3 0

2 years ago

On a piece of paper, graph yz -2x - 2. Then determine which answer choice

alukav5142 [94]

Answer:

Step-by-step explanation:

y≥-2x-2

put x=0,y=0

0≥-2

which is true.

so (0,0) lies on it.

graph is a solid line.

so B satisfies it.

3 0

3 years ago

Factor completely 6x - 18.<br> 6(x + 3)<br> 6(x-3)<br> 6X (-18)<br> Prime

8090 [49]

Answer:

6(x-3)

Step-by-step explanation:

the common number for 6 and 18 is 6 so if you extract that from the expression then it turns to 6(x-3) which cannot be factored further

5 0

3 years ago

What two numbers multiply to 11 and add to -16

VladimirAG [237]

They're irrational numbers, so they can't be exactly written down.
Rounded to the nearest thousandth, they are

- 15.280
and
- 0.720 .

5 0

3 years ago

Read 2 more answers

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

3 years ago

Read 2 more answers