Ask question

Ask question

All categories

Nutka1998 [239]

3 years ago

11

You're given two side lengths of 6 centimeters and 9 centimeters. Which measurement can you use for the length of the third side

to construct a valid triangle? 3 centimeters 10 centimeters 12 centimeters 14 centimeters 18 centimeters

1 answer:

irina [24]3 years ago

6 0

Hey there! I've got the answer for you!

The answer is: A (3 centimeters) and B (10 centimeters)

I know these are correct! Hope this helps ^-^

Have a good day! (Would you mind if I got BRAINLIEST?

You might be interested in

3<br> 50%<br> Draw angle ABC of size 65º.<br> A-<br> B<br> X+

Snowcat [4.5K]

65* It is drawn with a round ruller...

5 0

3 years ago

What is the average rate of change for the line shown in the graph

kompoz [17]

Answer:

The average rate of change is 10.

Step-by-step explanation:

Since the average rate of change is the slope of a line, you'd solve this problem by finding the slope.

First, you can start by using the $\frac{y_2-y_1}{x_2-x_1}$ method.

Plug in the points (0,10) and (1,20).

$\frac{20-10}{1-0} =\frac{10}{1} =10$

Therefore your answer is 10.

Hope this helped!

6 0

2 years ago

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

5 0

3 years ago

A farm uses a circular system to water the crops. The arm on the system (the radius) is 400 meters long. What area of land can i

ElenaW [278]

To answer this question you would need to know the formula for the area of a circle, $\pi r^{2}$ . You would start by squaring 400, getting 160,000. You would then multiply 160,000 by 3.14, getting 502,400 meters squared as your answer.

8 0

4 years ago

Read 2 more answers

A company that produces an expensive stereo component is considering offering a warranty on the component. Suppose the populatio

Llana [10]

Answer:

Warranty of 66 months.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 84, \sigma = 9$

If the company wants no more than 2% of the components to wear out before they reach the warranty date, what number of months should be used for the warranty?

Only the lowest 2% will be replaced, so the warranty is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.

$Z = \frac{X - \mu}{\sigma}$

$-2.055 = \frac{X - 84}{9}$

$X - 84 = -2.055*9$

$X = 65.5$

Warranty of 66 months.

5 0

3 years ago