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WARRIOR [948]
3 years ago
10

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
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