Can you forma triangle with the sides of 5 inches 5 inches, and 5 inches?

Which of these strategies would eliminate a variable in the system of equations? x−2y=115x+3y=−11

mafiozo [28]

Answer:

multiply the top equation by -5, then add the equations

Step-by-step explanation:

5 0

4 years ago

Help question below please and thank you i will give brainliest

vodomira [7]

Answer: angle V = 41 degrees (choice B)

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Explanation:

We're told that RS = ST = RT. This means triangle RST has all three sides equal to one another. This triangle is equilateral. Each interior angle is 60 degrees. This is because each angle is y, and y+y+y = 180.

Since each angle of triangle RST is 60 degrees, this means angle STR is 60 degrees.

Let z be the measure of angle UTV. Note how

(angleSTR)+(angleUTV) = 180

60+z = 180

z = 180-60

z = 120

Making angle UTV to be 120 degrees

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Now we can focus on triangle UTV since we have expressions for each interior angle

U = 4x-9
T = 120
V = 5x+6

Add up those angles, set the sum equal to 180, solve for x

U+T+V = 180

(4x-9)+(120)+(5x+6) = 180

9x+117 = 180

9x = 180-117

9x = 63

x = 63/9

x = 7

Use that x value to find the missing angle measures

angle U = 4x-9 = 4*7-9 = 28-9 = 19 degrees

angle V = 5x+6 = 5*7+6 = 35+6 = 41 degrees

Note how

U+T+V = 19+120+41 = 180

which helps confirm our answer.

6 0

3 years ago

Read 2 more answers

This my last question, I need help

il63 [147K]

It would be. C (SAS)

8 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

nasty-shy [4]

Answer:

(9.6, 25.7) is a 80% confidence interval for the average net change in a student's score after completing the course.

Step-by-step explanation:

We have n = 6, $\bar{x} = 17.6667$ and s = 13.3367. The confidence interval is given by

$\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n-1=5 degrees of freedom. As we want the 80% confidence interval, we have that $\alpha = 0.2$ and the confidence interval is $17.6667\pm t_{0.1}(\frac{13.3367}{\sqrt{6}})$ where $t_{0.1}$ is the 10th quantile of the t distribution with 5 df, i.e., $t_{0.1} = -1.4759$ . Then, we have $17.6667\pm (1.4759)(\frac{13.3367}{\sqrt{6}})$ and the 80% confidence interval is given by (9.6, 25.7)