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2 years ago

Solve each inequality: 4p+3>9p+6-5p

Mathematics

1 answer:

slava [35]2 years ago

7 0

Answer: No solution

Step-by-step explanation: when you subtract 5p from 9p you get 4p. Now you have 4p on both sides. To even it out, you have to subtract 4p from 4p. Not you are only left with the 3 and 6. The final answer is 3>6 which isn't true because 3 is LESS than 6. That is why this question has no solution.

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8. A computer application generates a sequence of musical notes using the function f(n) 6(16)", where n is the

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Applying an exponential property, it is found that the function that will generate the same note sequence as function f(n) is given by:

B. $H(n) = 6(2)^{4n}$

<h3>What is function for the note sequence?</h3>

The function for the node sequence is defined by:

$f(n) = 6(16)^n$

A function that will the same note sequence as function f(n) has the same initial value of 6. Additionally, applying an exponential property, we have that:

$H(n) = 6(2)^{4n} = 6(2^4)^n = 6(16)^n$

Hence option B is correct.

More can be learned about exponential properties at brainly.com/question/25537936

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1 year 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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2 years ago