All categories

2 years ago

Which of the following is equal to 2,952 ÷ 24?

Mathematics

1 answer:

attashe74 [19]2 years ago

5 0

Answer:

Step-by-step explanation:

After doing long division we then know that 2,952 ÷24 = 123

We 1st follow pemdas knowing this we solve the equations in parenthesis 1st

(2,400 ÷ 24) + (480 ÷ 24) + (72 ÷ 24)

2,400 ÷ 24 = 100

480 ÷ 24 = 20

72 ÷ 24 = 3

We can then rewrite the equation as

100 + 20 + 3 We then solve left to right

100 + 20 = 120

120 + 3 = 123

You might be interested in

Simplify 2x+y-3^3+1/2=

tatiyna

The answer is 2x + y - 53/2

7 0

3 years ago

Two more than a certain number is 15 less than twice the number. Find the number

Svetllana [295]

2+x=2x-15
>> 2+15=2x-x
>> 17 = x

So the answer is <u>17</u>

6 0

3 years ago

Read 2 more answers

Evaluate \dfrac j4 4 j ? start fraction, j, divided by, 4, end fraction when j=12j=12j, equals, 12.

cricket20 [7]

We need to evaluate the value of fraction $\frac{j}{4}$ .

We also given j=12.

In order to evaluate the value of fraction $\frac{j}{4}$ , we need to plug j=12.

Plugging j=12, we get

$\frac{j}{4}= \frac{12}{4}$

We have 12 in numerator and 4 in denominator.

We always divide top number by bottom number.

So, we need to divide 12 by 4.

On dividing 12 by 4 we get 3.

<h3>Therefore, $\frac{j}{4} = 3.$ </h3>

3 0

3 years ago

Read 2 more answers

Lance baked a pie Justin ate 2/3 of the pie and Joey ate 1/4 of the pie how much of the pie is left

zepelin [54]

1 - 2/3 - 1/4
= 1 - 8/12 - 3/12
= 1/12

8 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago