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2 years ago

What is the volume of a box that will hold exactly 567 of these cubes with 1/3 inch sides?

Mathematics

1 answer:

natima [27]2 years ago

6 0

21-inch³ is the Answer

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Solve for the Simplest form

tino4ka555 [31]

How to solve:

$\frac{5}{6}$ - $\frac{3}{8}$

Write all numerators above the least common denominator which is 24 in this case.
$\frac{20-9}{24}$

Subtract the numbers
$\frac{11}{24}$

Alternate form: 0.458333

The final, simplified, fraction answer is: $\frac{11}{24}$

7 0

3 years ago

Read 2 more answers

DO ALL QUESTIONS FOR BRAINLIST!! ( no links or files and make sure it’s right)

Slav-nsk [51]

Answer:

answers below

Step-by-step explanation:

1)5000

2)6450

3)50

4)30

5)1500

please make brainliest hope this helps

8 0

2 years ago

Smith cut his peanut butter

Lorico [155]

The answer is the square root of 101.75
Assuming that the sandwich Is a square, after cutting it diagonally, it turns into 2 triangles. To look for a missing length of a triangle, you use Pythagorean’s theorem a^2+b^2=c^2
Since c is 12 and a is 6.5 you just square 12 and square 6.5
144 42.25
Subtract it
101.75
You then have to find the square root of it. You can leave it as it is as √101.75 or just solve for it and round it to 10.09

3 0

2 years ago

A baseball bat factory produces 24,000 bats per day and uses 50 boxes for packing the bats. If the number of boxes needed for pa

Kisachek [45]

Answer:

Total boxes= 75

Step-by-step explanation:

Giving the following information:

A baseball bat factory produces 24,000 bats per day and uses 50 boxes for packing the bats.

<u>First, we need to determine the number of bats that fits into a box:</u>

Bats per box= 24,000 / 50

Barts per box= 480

<u>Now, the number of boxes for 36,000 bats:</u>

Total boxes= total bats / bats per box

Total boxes= 36,000/480

Total boxes= 75

3 0

2 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago