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3 years ago

Find the prime factorization of 504

Mathematics

1 answer:

BaLLatris [955]3 years ago

4 0

It's not a prime number but
Answer : 23 • 32 • 7

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<img src="https://tex.z-dn.net/?f=%283x3%29%2B%283x5%29-%282x-2%29" id="TexFormula1" title="(3x3)+(3x5)-(2x-2)" alt="(3x3)+(3x5)

MAVERICK [17]

We can simplify this problem by pairing like terms together.

9x + 15x - 2x + 2
22x + 2

This expression can be simplified down to
22x + 2

8 0

2 years ago

Read 2 more answers

Jefferson school has students in 1st grade up to 5st grade. The number of children in 1st grade has 3 digits.The digit in the nu

lapo4ka [179]

Answer:

283 or 382

Step-by-step explanation:

The number of children in 1st grade has 3 digits.The digit in the number are 2,3,and 8 .

Since the digit 8 means 80 in the number. The other two numbers could interchange the Hundred and Unit position respectively.

The possible numbers are:

I. 200+80+3=283

II. 300+80+2=382

The possible number of children in 1st Grade is either 283 or 382.

7 0

2 years ago

Read 2 more answers

The answer Is no A<br>please help

Alenkasestr [34]

Answer:

the answer is d

\can i get a brainliest pls

4 0

3 years ago

The area of a cross section perpendicular to the base of a rectangular prism is 45 square inches. If the length and width of the

Naily [24]

The height of the prism is $2.25 \ in$

Explanation:

It is given that the area of the cross section of the rectangular prism is 45 square inches.

Length of the rectangular prism is $5 \ in$

Width of the rectangular prism is $5 \ in$

To determine the height of the prism, let us substitute these values in the formula $SA=(2l+2w)h$ , we get,

$45=(2(5)+2(5))h$

Multiplying the terms within the bracket, we get,

$45=(10+10)h$

Adding the terms within the bracket, we have,

$45=20h$

Dividing both sides by 20, we get,

$2.25=h$

Thus, the height of the rectangular prism is $2.25 \ in$

4 0

3 years ago

Find the general solution of (x+3)y’=2y

gregori [183]

Answer:

$y=C(x+3)^2$

Step-by-step explanation:

We are given:

$\displaystyle (x+3)y^\prime=2y$

Separation of Variables:

$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}$

So:

$\displaystyle \frac{dy}{y}=\frac{2}{x+3} \, dx$

Integrate:

$\displaystyle \int\frac{dy}{y}=\int\frac{2}{x+3}\, dx$

Integrate:

$\displaystyle \ln|y|=2\ln|x+3|+C$

Raise both sides to e:

$|y|=e^{2\ln|x+3|+C}$

Simplify:

$|y|=(e^{\ln|x+3|})^2\cdot e^C$

So:

$|y|=C|x+3|^2$

Simplify:

$y=\pm C(x+3)^2=C(x+3)^2$

5 0

3 years ago