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2 years ago

6

Is n=3 a solution?-21 =

}{n}" alt="\frac{-63}{n}" align="absmiddle" class="latex-formula">

2 answers:

nadezda [96]2 years ago

7 0

Yes. -63 / 3 = -21.

Please consider branliest! ^_^

Orlov [11]2 years ago

5 0

Answer:

Yes it is, -63 / 3 = <u><em>-21</em></u>.

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Two people independently each pick a number from 1 to 10. what is the probability that the two of them pick the same number?

Marysya12 [62]

1 in 10 is the probability of two people picking the same number.

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3 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

Novay_Z [31]

Answer:

18225

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Consider the following division of polynomials.

Bond [772]

$x^4=x^2\cdot x^2$ . Multiplying the denominator by $x^2$ gives

$x^2(x^2+2x+8)=x^4+2x^3+8x^2$

Subtracting this from the numerator gives a remainder of

$(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8$

$-x^3=-x\cdot x^2$ . Multiplying the denominator by $-x$ gives

$-x(x^2+2x+8)=-x^3-2x^2-8x$

and subtracting this from the previous remainder gives a new remainder of

$(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8$

This last remainder is exactly the same as the denominator, so $x^2+2x+8$ divides through it exactly and leaves us with 1.

What we showed here is that

$\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}$

$=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}$

$=x^2-x+1$

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

$(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)$

$=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)$

$=x^4+x^3+7x^2-6x+8$

which matches the original numerator.

3 0

2 years ago

Read 2 more answers

NEED HELP ASAP!!!!!

Law Incorporation [45]

Answer:the length 7

Step-by-step explanation:each one is times 3 6x3 8x3 3x7=21

3 0

1 year ago

Given the Homogeneous equation x^2ydy+xy^2dx=0, use y=ux, u=y/x and dy=udx+xdu to solve the differential equation. Solve for y.

scZoUnD [109]

Answer:

$y=\frac{C}{x}$ .

Step-by-step explanation:

Given homogeneous equation

$x^2ydy+xy^2dx=0$

$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{xy^2}{x^2y}$

Substitute y=ux , $u=\frac{y}{x}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{y}{x}$

Now,

$u+x\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}x}$

$u+x\frac{\mathrm{d}u}{\mathrm{d}x}=-u$

$\frac{\mathrm{d}u}{\mathrm{d}x}=-2u$

$\frac{du}{u}=-\frac{dx}{x}$

Integrating both side we get

lnu=-2lnx+lnC

Where lnC= integration constant

$lnu+ln{x}^2=lnC$

$lnux^2=lnC$

Cancel ln on both side

$ux^2=C$

Substitute $u=\frac{y}{x}$

Then we get

xy=C

$y=\frac{C}{x}$ .

Answer: $y=\frac{C}{x}$ .

8 0

2 years ago