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2 years ago

Determine the Prime factorization of each number. 36 b. 120 c. 258 d. 63

Mathematics

2 answers:

Snezhnost [94]2 years ago

5 0

Answer:

the answer to this question is 36

Paha777 [63]2 years ago

5 0

Answer:

Step-by-step explanation:

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Josiah invests $360 into an account that accrues 3% interest annually. Assuming no deposits or withdrawals are made, which equat

Natalka [10]

y = 360(1.03)x is the answer

6 0

2 years ago

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How many zero arin the product of any whole. number less than 10 and 500

Stolb23 [73]

Answer:

Hay un máximo de tres ceros en el producto de un número distinto de cero cero menor que 10 y 500

Para poder ver esto, la forma más fácil para resolver este problema es multiplicar todos los números entre 1 y 9 por 500, es decir:

01*500 = 500

2*500 = 1.000

3*500 = 1.500

4*500 = 2.000

5*500 = 2.500

6*500 = 3.000

7*500 = 3.500

8*500 = 4.000

9*500 = 4.500

Como vemos, si multiplicamos a 500 por cualquier número par, entonces obtenemos un número con tres ceros, mientras que si este es impar solamente obtenemos dos ceros

dame coronaa

7 0

3 years ago

Read 2 more answers

The sum of 13 divided by a number and that number divided by 13

7nadin3 [17]

Sum means add

the number is x

13 divided by a number is 13/x
the number divided by the number is x/13

so
$\frac{13}{x}+\frac{x}{13}$
also can be simplified to
$\frac{169}{13x}+\frac{x^2}{13x}$ or
$\frac{x62+169}{13x}$

translated it is $\frac{13}{x}+\frac{x}{13}$ where x is the number

5 0

2 years ago

HELP ME PLZ. I have no clue how to do this.

Varvara68 [4.7K]

Answer:

Step-by-step explanation:

3x+1=8x-24

25=5x

x=5

3 times 5 plus 1 is 16 and so the length is 32

comment section:

XM=MY because MN is segment bisector

XM=3x+1=3*5+1=16=MY and since XM+MY=XY 16+16=32

3 0

2 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago