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suter [353]
3 years ago
9

In a rectangle KLMN, KM = 6x + 16 and LN = 49. find the value of x

Mathematics
2 answers:
RoseWind [281]3 years ago
5 0

Answer:

5.5

Step-by-step explanation:

in a rectangle, opp. sides are equal,

∴, KM=LN

∴, 6x+16=49

6x = 33

x = 5.5


balu736 [363]3 years ago
5 0

Answer:

5.5

Step-by-step explanation:

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X − 3y if x = 3 and y = −2.
Korvikt [17]

Answer:

Step-by-step explanation:

x - 3y

3 - 3(-2) = 3 + 6 = 9

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3 years ago
First, rewrite 9/10 and 23/25 so that they have a common denominator.
MAVERICK [17]
The cheap answer is, well, all we do is, grab the denominator of one and multiply the other by it, top and bottom, and grab the denominator of the other, and multiply the first one by that one too, that way, both will have the same denominator, and then you can simply check the numerator to see who's larger, let's do so.

\bf \cfrac{9}{\boxed{10}}\qquad \qquad \cfrac{23}{\boxed{25}}
\\\\\\
\cfrac{9\cdot \boxed{25}}{10\cdot \boxed{25}}\implies \cfrac{225}{250}\qquad \qquad \qquad \cfrac{23\boxed{10}}{25\cdot \boxed{10}}\implies \cfrac{230}{250}\\\\
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\textit{so, which is larger then?}\qquad \cfrac{225}{250}~or~\cfrac{230}{250}?

surely you can tell.
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3 years ago
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Novay_Z [31]

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Step-by-step explanation:

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3 years ago
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Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

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Answer:

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