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Paraphin [41]
4 years ago
12

(see pic) please help, and try to explain, I forgot how to do them

Mathematics
1 answer:
aleksley [76]4 years ago
7 0

Answer:

  (D) I and II only

Step-by-step explanation:

A function has a limit at a point if the limit from the left and the limit from the right are the same value.

Here f(3) = 5, and the linear function defined for x ≤ 3 is continuous, so the left limit exists.

The function defined for x > 3, when evaluated at x = 3, also has a value of 5. That function, too, is continuous up to x = 3, so the right limit is defined.

The left and right limits at x=3 are y=5, so the limit exists. (Statement I is true.)

__

The function f(x) is defined as 5 for x=3, so the limit exists and is the same as the function value at x=3. That means the function is continuous at x=3. (You can draw the graph through that point without lifting your pencil.) (Statement II is true.)

__

The slope of f(x) for x < 3 is 1; the slope for x > 3 is 4. There is a discontinuity in the slope at x=3, so the function is not differentiable there. (Statement III is false.)

_____

The graph shows the function in red and its derivative in blue. You will notice there is a discontinuity (step change) at x=3.

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Answer:

Option A.

Step-by-step explanation:

Let M be the milk per gallon.

C be the cookies per dozen.

B be the bundle (one gallon of milk and a dozen cookies ).

Milk can be sold by itself for a profit of $1.50 per gallon. Cookies can likewise be sold at a profit of $2.50 per dozen and bundle is sold for a profit of $3.00 per bundle.

Profit= 1.5M + 2.5C + 3B

We need to maximize the profits. So, our objective function is

Therefore, the correct option is A.

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Answer:

y = 3

Step-by-step explanation:

8(y + 2) = 40 \\ y + 2 =  \frac{40}{8}  \\ y + 2 = 5 \\ y = 5 - 2 \\ y = 3

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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
3 years ago
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