Evaluate each expression by replacing the variable with 8. Show each step. (8 marks) 2n - 6

Mathematics

1 answer:

monitta3 years ago

4 0

Answer:

Step-by-step explanation:

2n-6

n=8

2(8)-6

16-6

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Solve the following inequality algebraically.<br> x + 2 < 11

Pachacha [2.7K]

Answer:

x < 9

Step-by-step explanation:

Subtract 2 from both sides so you can isolate the variable on the left.

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2 years ago

For the region bounded by y=2x, the x-axis, and x=1, determine which of the following is greater: the volume of the solid ge

telo118 [61]

Answer:

The correct answer B) The volumes are equal.

Step-by-step explanation:

The area of a disk of revolution at any x about the x- axis is πy² where y=2x. If we integrate this area on the given range of values of x from x=0 to x=1 , we will get the volume of revolution about the x-axis, which here equals,

$\int\limits^1_0 {(pi)y^{2} } \, dx$

which when evaluated gives 4pi/3.

Now we have to calculate the volume of revolution about the y-axis. For that we have to first see by drawing the diagram that the area of the CD like disk centered about the y-axis for any y, as we rotate the triangular area given in the question would be pi - pi*x². if we integrate this area over the range of value of y that is from y=0 to y=2 , we will obtain the volume of revolution about the y-axis, which is given by,

$\int\limits^2_0 {pi - pix^{2} } \, dy = \int\limits^2_0 {1 - y^{2}/4 } \, dy$

If we just evaluate the integral as usual we will get 4pi/3 again(In the second step i have just replaced x with y/2 as given by the equation of the line), which is the same answer we got for the volume of revolution about the x-axis. Which means that the answer B) is correct.

7 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$