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3 years ago

If JO = 32, is triangle JKO is congruent PMN to triangle ? If so, find JN.

1 answer:

8 0

It is true that they are congruent.
Side JK and PM are congruent to each other
JK = 25
PM is divided into two measurements=19 and 6.
So PM is 19 + 6 = 25
Therefore JK = PM

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Marcel is purchasing a board to build a bookcase. He wants to divide the board into 1.75 foot sections . He need six sections .

FromTheMoon [43]

10.5 feet (1.75x6) hope I helped

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3 years ago

3000 exercise books are arranged into 3 piles. The first pile has 10 more books than the second pile. The number of books in the

pochemuha

Answer:

598 books

Step-by-step explanation:

Let x be the number of books in the 3rd pile. We will have then:

1st pile: 2x + 10

2nd pile: 2x

3rd pile: x

3000 = (2x + 10) + (2x) + (x)

3000 = 5x + 10

3000 - 10 = 5x

2990 = 5x

2990/5 = x

x = 598 books in the third pile!

We can check that:

3000 = 2*598 + 10 + 2*598 + 598

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3 years ago

Rearrange the following equation to solve for x. y=1/2x+10

maxonik [38]

In order to solve this we must first,

Subtract both sides of the equation by 10,
y-10=1/2x

Now we have to multiply both sides by 2 so that only x is left.

2y-20=x

Therefore, x=2y-20

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3 years ago

Y-x=2x-2/3y <br> Written in y=mx+b form

FromTheMoon [43]

1 2/3y = 3x
A picture is worth a thousand words

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3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

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3 years ago