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kupik [55]
3 years ago
8

What is the difference?

Mathematics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer:

<h2>D. \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\ or StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction</h2>

Step-by-step explanation:

Given the expression \frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }, the dfference is expressed as follows;

Step1: First we need to factorize the denominator of each function.

\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }\\= \frac{x}{x^{2}-5x+3x-15 } - \frac{4}{x^{2} + 7x-5x - 35 }\\= \frac{x}{x(x-5)+3(x-5) } - \frac{4}{x( x+ 7)x-5(x +7) }\\= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\\\

Step 2: We will find the LCM of the resulting expression

=  \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\= \frac{x(x+7)-4(x+3)}{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+7x-4x-12 }{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\

The final expression gives the difference

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A pan is heated to 393°F, then removed from the heat and allowed to cool in a kitchen where the room temperature is a constant 6
deff fn [24]

Answer:

The approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

Step-by-step explanation:

The formula for D, the difference in temperature between the pan and the room after t minutes is:

D = 324\cdot e^{-0.05t}

Compute the approximate difference in temperature between the pan and the room after 9 minutes as follows:

D = 324\cdot e^{-0.05t}

    =324\times e^{-0.05\times 9}\\\\=206.59

Then the approximate temperature of the pan after it has been away from the heat for 9 minutes is:

D = P - R

206.59 = P - 69

P = 206.59 + 69

P = 275.59°F

Thus, the approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

6 0
3 years ago
For the function h(x) defined below, find function f(x) and g(x) so that h(x)=(f^ g)(x) . h(x) = ((x + 2)/x) ^ 3
dem82 [27]

Answer:

Step-by-step explanation:

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4 0
3 years ago
How do you do this question?
Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\

\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

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Answer:

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Step-by-step explanation:

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27×⅓

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