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3 years ago

What's a number greater than 40 whose prime factorization contains 3 prime numbers that do not repeat

1 answer:

3 0

Well to make it easy pick 3 prime numbers. In this case I'll use 3, 11, and 13. Multiply all 3 together to get 429. For the second part, pick a prime number and multiply it by 3, in this case 3, which when multiplied by 3 equals 9. Multiply that by a prime number squared, like 11 which is 121. Then multiply by 13. 3x3x3x11x11x13 = 42471 THE CREDIT GOES TO ICEDRAPTOR

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Can someone please help and explain this

nadezda [96]

Using proportional
x/10 = 14/8
x = 10*14/8
x = 17.5

answer
C. x =17.5

6 0

2 years ago

Read 2 more answers

Simplify (64)3<br> O A. 69<br> O B. 612<br> O c. 63<br> O d. b

Neko [114]

The answer will be 192.

8 0

2 years ago

Read 2 more answers

Betty plans to have two children but doesn't know if they will be boy-boy, girl-girl, or boy-girl. What is the probability that

Doss [256]

50% every birth has a 50 50 chance of boy or girl

6 0

3 years ago

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from

Maslowich

I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for $C$ . This is easy enough to do. First fix any one variable. For convenience, choose $x$ .

Now, $x^2=2y\implies y=\dfrac{x^2}2$ , and $3z=xy\implies z=\dfrac{x^3}6$ . The intersection is thus parameterized by the vector-valued function

$\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle$

where $0\le x\le 4$ . The arc length is computed with the integral

$\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx$

Some rewriting:

$\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}$

Complete the square to get

$x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4$

So in the integral, you can substitute $y=x^2+\dfrac92$ to get

$\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy$

Next substitute $y=\dfrac{\sqrt{63}}2\tan z$ , so that the integral becomes

$\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz$

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

$\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C$

So the arc length is

$\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98$

4 0

3 years ago

sally, sara and spike all made their own soaghetti sauce using a combination of exactly 50 tomatoes, onions, and cloves of garli

Ksju [112]

one could use 1 tomato 48 onions, and 1 clove of garlic

another could be 6 tomatos 40 onions and 4 cloves of garlic,

last could be 10 tomotos 10onions and 30 cloves of garlic

5 0

2 years ago