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3 years ago

Order the steps for Multiplying/Dividing Rational Expressions correctly.

1 answer:

6 0

We have been given some steps but not in order about how to simplify a rational expression. Now we just need to rearrange them in correct order so the correct order will be:

Step 1): For division, flip the second fraction (KCF).

Step 2): Factor the numerator and denominator completely.

Step 3): Divide out common factors (Math Ninja!).

Step 4): Multiply the simplified fractions.

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The angles formed by two perpendicular lines have a measure of _______.

Tpy6a [65]

The answer is 90 degrees. SO it would be B

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3 years ago

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Can somebody help me out ??

julia-pushkina [17]

Answer: y = -3/2x -5

explanation:
y = mx + b
7 = -3/2(-8)+b
7= 12 +b
subtract 12 on both sides
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y = -3/2x - 5

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3 years ago

a truck is carrying waterbottles. 1/4 of the the total number of bottles in the truck plus 18 bottles is equal to 56 bottles. ho

frosja888 [35]

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4 years ago

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There are 4 elephants in length for every blue whale, 4 blue whales for every

OlgaM077 [116]

Answer:

Step-by-step explanation:

4 elephants = 1 blue whale

4 blue whales = 1 tree = 16 elephants

2 trees = 1 monument = <u>32 elephants</u>

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3 years ago

onsider the equation below. (If an answer does not exist, enter DNE.) f(x) = 8 cos2(x) − 16 sin(x), 0 ≤ x ≤ 2π (a) Find the inte

MariettaO [177]

Answer:

(a) Increasing: $\frac{\pi}{2}< x< \frac{3\pi}{2}$ and Decreasing: $0< x< \frac{\pi}{2}\ \text{or}\ \frac{3\pi}{2}< x< 2\pi$

(b) The local minimum and maximum values are -16 and 16 respectively.

Step-by-step explanation:

The function provided is:

$f(x)=8cos^{2}(x)-16sin( x);\ 0\leq x\leq 2\pi$

(a)

$f(x)=8cos^{2}(x)-16sin( x);\ 0\leq x\leq 2\pi$

Then, $f'(x)=-16cos(x)sin(x)-16cos(x)=-16cos(x)[1+sin(x)]$

Note, $1+sin(x)\geq 0\ \text{and }\ sin(x)\geq 1\\$

Then, $sin(x)=-1\Rightarrow x=\frac{3\pi}{2}$ for $0\leq x\leq 2\pi$ .

Also $cos(x)=0$ .

Thus, f (x) is increasing for,

$f'(x)>0\\\Rightarrow cos(x)$

And f (x) is decreasing for,

$f'(x)0\\\Rightarrow 0< x< \frac{\pi}{2}\ \text{or}\ \frac{3\pi}{2}< x< 2\pi$

(b)

From part (a) f (x) changes from decreasing to increasing at $x=\frac{\pi}{2}$ and from increasing to decreasing at $x=\frac{3\pi}{2}$ .

The local minimum value is:

$f(\frac{\pi}{2})=8cos^{2}(\frac{\pi}{2})-16sin(\frac{\pi}{2})=-16$

The local maximum value is:

$f(\frac{3\pi}{2})=8cos^{2}(\frac{3\pi}{2})-16sin(\frac{3\pi}{2})=16$

(c)

Compute the value of f'' (x) as follows:

$f''(x)=16sin(x)[1+sin(x)]-16cos^{2}(x)\\\\=16sin(x)+16sin^{2}(x)-16[1-sin^{2}(x)]\\\\=32sin^{2}(x)+16sin(x)-16\\\\=16[2sin(x)-1][sin (x)+1]$

So,

$f''(x)>0\\\Rightarrow sin(x)>\frac{1}{2}\\\Rightarrow \frac{\pi}{6}$

And,

$f''(x)$

Thus, f (x) is concave upward on $(\frac{\pi}{6},\ \frac{5\pi}{6})$ and concave downward on $(0,\ \frac{\pi}{6}), (\frac{5\pi}{6},\ \frac{3\pi}{2})\ \text{and}\ (\frac{3\pi}{2},\ 2\pi)$ .

If $x=\frac{\pi}{6}$ , then f (x) will be:

$f(\frac{\pi}{6})=8cos^{2}(\frac{\pi}{6})-16sin(\frac{\pi}{6})=-2$

If $x=\frac{5\pi}{6}$ , then f (x) will be:

$f(\frac{5\pi}{6})=8cos^{2}(\frac{5\pi}{6})-16sin(\frac{5\pi}{6})=-2$

The inflection points are $(\frac{\pi}{6},\ -2)\ \text{and}\ (\frac{5\pi}{6},\ -2)$ .

7 0

4 years ago