All categories

3 years ago

An ordered set of numbers or objects is a

Mathematics

1 answer:

aniked [119]3 years ago

4 0

The answer is pattern

You might be interested in

Which set of reflections would carry rectangle ABCD onto itself?

Whitepunk [10]

Answer:

i think that the answer would be but im not 100% sure

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A cube has a surface area of 3/2. What's the volume?

Cloud [144]

Answer:

Step-by-step explanation:

Assume the surface area = 6x²

6 is the number of faces and x is the length of the cube

so 3/2 = 6x²

Divide by 6

1/4 = x²

Square root both sides

1/2 = x

Volume is : l×w×h

so 1/2 × 1/2 × 1/2 = 1/8

5 0

3 years ago

Read 2 more answers

If you use 104 kiloliters of water per year, approximately how many liters of water do you use in a day?

tatiyna

Answer:

0.28 liters

Step-by-step explanation:

104÷365=0.28

5 0

2 years ago

Read 2 more answers

A smoothie shop has 40 stores and 55% of the stores are in california the rest of the stores are in nevada.how many are in nevad

ZanzabumX [31]

There are 18 in Nevada.

Multiply 55% by 40, and you get the amount for California. (Which is 22.)

Subtract 40 by 22, and you get 18.

Hope this helps!

5 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago