1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
victus00 [196]
4 years ago
8

Find the GCF of the monomials: 72x^3 y^2 and 210x^2y^5

Mathematics
2 answers:
denpristay [2]4 years ago
4 0

Answer: The GCF of the monomials is 1.

Elden [556K]4 years ago
3 0

For this case we have by definition, that the Greatest Common Factor or GFC, of two or more whole numbers is the largest integer that divides them without leaving a residue.

Now, we look for the factors of 72 and 210:

72: 1,2,3,4,6,8,9,12,16,24,36,72

210: 1,2,3, 5,6,7 ...

Thus, the GFC of both is 6.

Then, the GFC of 72x ^ 3y ^ 2 and 210x ^ 2y ^ 5 is given by:

6x ^ 2y ^ 2

ANswer:

6x ^ 2y ^ 2

You might be interested in
Write an equation in point-slope form for the line that passes through the point with the given slope. (–2, 11), m=43
frosja888 [35]

Answer:

y-11=43(x+2)

Step-by-step explanation:

y-y1=m(x-x1)

y-11=43(x-(-2))

y-11=43(x+2)

4 0
3 years ago
(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
Nancy keeps the feed for the chickens in a large square bin. The base of the bin is square and has an area of 16 900 cm². What i
Nikolay [14]

Answer:

area= l×b or b^2 or l^2 = 16900

to get one side length of the square take root of area i.e

\sqrt{16900}

=130 cm is length of the base

4 0
3 years ago
Plot 2/25 on the number line.
S_A_V [24]
The answer would be .8
6 0
3 years ago
Read 2 more answers
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
Galina-37 [17]

Answer:

Step-by-step explanation:

Hello, we know that if the equation is

   y=a(x-h)^2+k

Then the vertex is the the point (h,k)

Here, the vertex is the point (-2,-20) so we can write, a being a real number that we will have to find,

y=a(x-(-2))^2-20=a(x+2)^2-20

On the other hand, we know that the y-intercept is (0,-12) so we can write

-20=a(0+2)^2-12=4a-12\\\\\text{We add 12 and we divide by 4.}\\\\4a = -20+12=-8\\\\a = \dfrac{-8}{4}=-2

So the equation becomes.

\boxed{y=-2(x+2)^2-12}

And we can give the standard form as below.

y=-2(x+2)^2-12=-2(x^2+4x+4)-12\\\\=-2x^2-8x-8-12 \ \\\\\boxed{y=-2x^2-8x-20}

Thank you.

7 0
4 years ago
Other questions:
  • Valentino's Pizzeria sells four types of pizza crust. Last week, the owner tracked the number sold of each type, and this is wha
    13·1 answer
  • A coin is tossed three times the probability of zero heads is 1/8 what is the probability that all three tosses result in the sa
    7·1 answer
  • Identify the vertex, focus, and directrix of the parabola with the equation x^2-6x-8y+49=0
    10·2 answers
  • Which are solutions of the equation (x + 5)(x – 3) = 0? Check all that apply.
    7·2 answers
  • Can anyone help me ???
    14·1 answer
  • The perimeter of a square field is 328 yards. How long is each side?
    8·2 answers
  • Change 34% to a decimal
    14·1 answer
  • There are 12 pupils in a class
    14·1 answer
  • I'm stuck on this question
    7·2 answers
  • What is the slope of 2/7x +5y=15
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!