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3 years ago

Write the prime factorization of 500 using exponents

Mathematics

1 answer:

murzikaleks [220]3 years ago

6 0

Answer :Ex.: 2, 3, 4, 11, 10225, etc

Step-by-step explanation:

The number 500 is a composite number so, it is possible to factorize it. In other words, 500 can be divided by 1, by itself and at least by 2 and 5. A composite number is a positive integer that has at least one positive divisor other than one or the number itself. In other words, a composite number is any integer greater than one that is not a prime number.

The prime factorization of 500 = 22•53.

The prime factors of 500 are 2 and 5.

Factor tree or prime decomposition for 500

As 500 is a composite number, we can draw its factor tree:

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Can someone tell me the answers to the questions i just imputed? :>

Andrej [43]

It is 54 for 1 and 2 is 2520 you multply the lenght by the whith and divide by 2

6 0

3 years ago

I need help, i get like 4712, but its not an answer

ella [17]

So you will need to find the height, and we can do so by using the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ . Using this formula, we can make the equation $15^2+h^2=25^2$

Firstly, solve the exponents: $225+h^2=625$

Next, subtract 225 on both sides: $h^2=400$

Lastly, square root both sides, and your answer is h = 20.

Now that we got the height, we can solve for the volume, which is $V=\pi r^2\frac{h}{3}$ . Using this formula, we can form our equation to be $V=\pi 15^2\frac{20}{3}$

Firstly, solve the exponent: $V=\pi*225*\frac{20}{3}$

Next, multiply 225 and 20/3, and your answer will be V = 1500π, or the third option.

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.