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1 year ago

Which of the following statement(s) describes the rate of change of f over the interval 1.5 ≤ x ≤ 3? Select all that apply.

Mathematics

1 answer:

Vesnalui [34]1 year ago

6 0

2 Answers: Choice B and Choice C

The rate of change is 2.

The rate of change is constant.

======================================================

Explanation:

The first point on the left is when x = 1.5 and it has a height of y = 1

The point (1.5, 1) is on the line.

So is the point (3,4) for similar reasoning.

Compute the slope between those points

m = (y2-y1)/(x2-x1)

m = (4-1)/(3-1.5)

m = 3/(1.5)

m = 2

The slope is 2, which is the same as saying the rate of change is 2. This only applies when x > 1 of which the interval 1.5 ≤ x ≤ 3 is a part of.

Since the slope stays at 2 on the interval 1.5 ≤ x ≤ 3, this means we consider the slope to be constant. If the curve bended at all on this interval, then it wouldn't be a constant slope.

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3 years ago

Twice the sum of consecutive numbers n and n + 1 is 510. What is the smaller number?

guajiro [1.7K]

Answer:

The smaller number is 127

Step-by-step explanation:

Lets write the given problem in equation form

sum of consecutive numbers n and n + 1 = n + n + 1 = 2n + 1

now we find twice of the sum of consecutive numbers n and n + 1

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thus,

4n + 2 = 510

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2 years ago

Can anyone help me integrate :

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Rewrite the second factor in the numerator as

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$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
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Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
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We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

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