Ask question

Ask question

All categories

3 years ago

6

Divide $54,860 by 34

2 answers:

Vinvika [58]3 years ago

6 0

Go on calculator type 54860 by 34 and see what the answer is

Anika [276]3 years ago

6 0

Answer:

$1,613.53

Step-by-step explanation:

You might be interested in

Easy points for those who are good at math!

zalisa [80]

<u>Answer:</u>

The answer is A) y = -7x - 7.

<u>Step-by-step explanation:</u>

First, you need to subtract '7x' from both sides, and once that is done, the equation is in slope-intercept form because it is in y=mx+b form. I'll attach a photo below that explains my process! :) And another possible answer would be y = 7x + 7, but that is not an option listed above.

<em>Hope this helps! Feel free to give me Brainliest if you feel this helped. Have a good day, and good luck on your assignment. :)</em>

4 0

2 years ago

The moment of intertia about the z-axis of the solid shown on the right with density deltaδequals=1 is upper i subscript z base

Ratling [72]

<span>私の名前はあなたのビジネスのどれでもありません</span>

8 0

3 years ago

BRAINLIEST ANSWER PLUS EXTRA POINTS! Two planes leaves an airport at the same time, one flying east one flying west. The eastbou

nirvana33 [79]

Equation:

distance + distance = 2250 miles

3x + 3(x-110) = 2250

6x - 330 = 2250

6x = 2580

x = 430 mph (rate of the westbound plane)

x-110 = 320 mph (rate of the eastbound plane)

PLS MARK BRAINLIEST!!!

5 0

3 years ago

You are playing a card game. if you pull an ace you win $26, if you pull the king of hearts you win $204, if you pull a jack or

andreyandreev [35.5K]

How much money do you lose if you don’t get either of those options?
Specify?

4 0

3 years ago

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

2 years ago

Read 2 more answers