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2 years ago

What is the product in lowest terms?

1 answer:

4 0

-5/12 × 8/13

First, we need to start out by applying the multiplication rule towards fractions. If you haven't learned or don't remember the rules, you can always look up fraction rules. The rule is: a/b × c/d = ac/bd. Basically, we are combining the numerators and both of the denominators.
$- \frac{5 \times 8}{12 \times 13}$

Second, let's now multiply what we have in the fraction. (5 × 8 = 40) and (12 × 13 = 156). Doing so will create a new fraction for us to use.
$-\frac{40}{156}$

Third, now obviously, we can simplify the fraction we just got into lower terms. To do that, we have to collect the greatest common factor (GCF) of both the numerator and denominator and list their factors to find the common factor that is the greatest.

Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40
Factors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156

Out of the listed factors, which of those are the common ones? 1, 2, and 4 are the common factors. Since we are looking for the greatest common factor, it would be 4 since that is the higher number out of the commons. The GCF is 4.

Fourth, we can now divide our numerator (40) and our denominator (156) by the GCF we just found out, which was 4.
$40 \div 4 = 10 \\ 156 \div 4 = 39$

Fifth, our last step is to create our new simplified fraction. All we have to do is take our new numerator and denominator to see the fraction.

Answer in fraction form: $\fbox {-10/39}$
Answer in decimal form: $\fbox {-0.2564}$

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Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$ , concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$ .

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3 years ago

Distance: (-2,7) and (2,-5)

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Does it say anything about slope?

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3 years ago

(The last 22 and 23)

quester [9]

1) 24
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Hope this helps

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3 years ago

Amal and Salma share some money in the

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$315, i cant give a thorough explanation since i did it all in my head but basically amal gets 5 portions of the money and Salma gets 9, meaning she gets 4 more, so if 4 portions is $252 then it cant be any of the others since they are too low or way too high, $315 is the only reasonable one.

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3 years ago

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A number is greater than 8. The same number is less than 10. The inequalities x greater-than 8 and x less-than 10 represent the

jok3333 [9.3K]

Answer:

There are a few solutions because there are some fractions and decimals between 8 and 10

Step-by-step explanation:

Let the unknown number be 'x'

If the number is greater than 8 and the same number is less than 10, this can be expressed as;

x>8 and x < 10

Note that if x>8, then 8<x

The resulting inequalities are now;

8<x and x<10

Combining both inequalities we have: 8<x<10

Since the inequality didn't tell us that the variable 'x' is equal to 8 and 10, this means that our solution falls between 8 and 10 and the value of integer that falls within this range is 9. Other values that falls within this range are decimals and fractions.

Therefore it can be concluded that there are a few solutions because there are some fractions and decimals between 8 and 10

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3 years ago

Read 2 more answers