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Dennis_Churaev [7]
3 years ago
12

What type of function does this scatter plot (graph A) look like?

Mathematics
1 answer:
NARA [144]3 years ago
7 0
I think the correct answer is linear
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A = {a, b, c, d}
Rama09 [41]

Answer:

No, Johnny is not correct

A U B = {a,b,c,d,e,f,g,h}

Step-by-step explanation:

Johnny's answer does not have c and d, elements that are in A and B, so A U B must have all elements that belong to A or B.

The (U) represents the sum of elements between two sets. So If a set has three elements {@,#,$} and another set has 5 elements {%,^,&}, (U) equals the sum of these sets: {@,#,$,%,^,&}.

The same is true for the sets A and B.

A = {a, b, c, d}

B = {c, d, e, f, g, h}

A U B = {a,b,c,d,e,f,g,h}

Note: c, d are repeated, in A and B. It is not necessary to write them twice in the final answer.

5 0
3 years ago
The table shows the monthly rainfall amount in Honolulu, Hawaii during one year.
tensa zangetsu [6.8K]

Answer:

Option D: 1.50

Step-by-step explanation:

8 0
2 years ago
6x + 5y =20 solve for y step by step
Marina CMI [18]

Answer:

Move all terms that don't contain y to the right side and solve.y=4−6x/5

Step-by-step explanation:

4 0
2 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}}

3 0
2 years ago
If the given figure is rotated 90 degrees counterclockwise around the origin, what are
Lisa [10]

Answer:

the answer is (-4,2)

hope this helps

7 0
2 years ago
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