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Find the radius of the circle. please.

labwork [276]

Answer:

12in

Step-by-step explanation:

2r = 24

r = 12

8 0

3 years ago

Read 2 more answers

You randomly choose one of the following numbers shown. Find the probability the event can occur choosing a number greater than

sergey [27]

<h3>Answer: 7/11</h3>

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Explanation:

The event space is the set of outcomes we want to happen.

The event space is {5, 6, 7, 8, 9, 10, 11} which consists of getting a value greater than 4. There are 7 items in this event space. Let A = 7.

The sample space consists of all possible outcomes. The sample space is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} and there are 11 items in the sample space. Let B = 11.

Divide the values of A and B to get the answer we're after

A/B = 7/11

This is the probability of selecting a value greater than 4.

7 0

3 years ago

The sum of three consecutive numbers is 153. What is the smallest of the three numbers ?

astra-53 [7]

$n;\ n+1;\ n+2 - \text{three consecutive numbers}\\\\(n)+(n+1)+(n+2)=153\\\\n+n+1+n+2=153\\\\3n+3=153\ \ \ \ |-3\\\\3n=150\ \ \ \ |:3\\\\n=50$

Answer: 50

6 0

4 years ago

Graph on a coordinate plane or explain how to graph the equation: y = 3x – 5

Vinvika [58]

Answer:

Step-by-step explanation:

Just substitute any values of x so that you can graph it <33

6 0

2 years ago

Identify the lines that are perpendicular: A. = −5 and = 2 are perpendicularB. + 5 = 2 and − 5 + = 3 are perpendicularC. = 13 +

scZoUnD [109]

In order to check if the lines are perpendicular, we need to check if their slopes have the following relation (to find the slope we can use the slope-intercept form y = mx + b):

$m_1=-\frac{1}{m_2}$

A.

In this option, y = -5 is an horizontal line and x = 2 is a vertical line, therefore they are perpendicular.

B.

First let's find the slope of each line:

$\begin{gathered} x+\frac{y}{5}=2 \\ 5x+y=10 \\ y=-5x+10\to m=-5 \\ \\ -\frac{x}{5}+y=3 \\ y=\frac{x}{5}+3\to m=\frac{1}{5} \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

C.

$\begin{gathered} y=\frac{1}{3}x+1\to m=\frac{1}{3} \\ \\ y-1=-3(x-5) \\ y-1=-3x+15 \\ y=-3x+16\to m=-3 \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

D.

$\begin{gathered} y-5=x+1 \\ y=x+6\to m=1 \\ \\ x+y=3 \\ y=-x+3\to m=-1 \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

E.

$\begin{gathered} y+2=\frac{1}{3}(x-6) \\ y+2=\frac{1}{3}x-2 \\ y=\frac{1}{3}x-4\to m=\frac{1}{3} \\ \\ y=3x+4\to m=3 \end{gathered}$

These slopes don't obey the relation stated above, so the lines aren't perpendicular.

The correct options are A, B, C and D.

5 0

2 years ago

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