What is the slope of a line perpendicular to the line whose equation is x + 3 y = − 3 . Fully reduce your answer.

Answers

Answer 1
Answer: X+3y=-3
3y=-x-3
y=-1/3x-1
So the slope of this line is -1/3.
Therefore, the slope of a perpendicular line is the negative reciprocal so it would be 3.
Answer 2
Answer:

Final answer:

In summary, a line is perpendicular to x + 3y = -3 if its slope is 3.

Explanation:

The subject requested is about finding the slope of a line that is perpendicular to the line specified by the equation x + 3y = -3. To find the slope, we'll first have to rewrite the given equation in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

Rewriting x + 3y = -3, we get y = -x/3 - 1. So the slope of the given line is -1/3. Now, two lines are perpendicular if and only if the product of their slopes is -1. Therefore, the slope of the line perpendicular to the given line will be -1 / original slope = -1 / (-1/3) = 3.

Thus, the slope of a line perpendicular to the line x + 3y = -3 is 3.

Learn more about Slope here:

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The unit rate of $1.12 for 8.2 ounces

Answers

That is around 13 or 14 cents. The specific answer is 13.65853659 cents, but you can either round up to 14 or round down to 13 (rounding up to 14 is the correct way to do it.)

First you divide 1.12 by 8.2 to get the unit rate because It takes $1.12 to buy 8.2 ounces. 1.12/8.2=13.65853659

Hope this helped!

How can you identify the parts of an expression using mathematical terms?

Answers

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5/25 = 13/?

Can you please help me with this question.

Answers

This a proportion.

Lets solve by cross multiplying.

5/25 = 13/x

5 * x = 5x                13 * 25 = 325

325 = 5x

5x/5 = 325/5
x= 325/5             x= 65

Answer:   5/25 = 13/65
5/25
This can be simplified to 1/5
1/5 = 13/?
In order to get to 13 from 1 you must multiply by 13
If we multiply the bottom (denominator) by 13 we get the missing value
5 x 13 = 65

5/25 = 13/65

Trampoline Park has an admission fee, plus an hourly fee of $4.50. Determine the admission fee that the trampoline park charges.What is the initial value?
A. $4.50
B. $5.00
C. $5.50
D. $6.00

Answers

Answer:

5.00 (or B)

Step-by-step explanation:

I just took it on edge and got it

trust me brother

!HELP! SHOW WORK PLEASE SO I CAN UDNERSTAND!A paper drinking cup in the shape of a cone has a height of 10 centimeters and a diameter of 8 centimeters. Which of the following is the closest to the volume of the cup in cubic centimeters?
167 cm³
209 cm³
670 cm³
502 cm³

Answers

Ok.

The volume of a cone is the volume of a cylinder over 3.

The volume of a cone is:
(1)/(3) \pi r^(2)h

This problem gives us the diameter, which is twice of the radius.  So, we'll just half it.
Radius (me) = 4 cm.

Substitute our given values into the equation.
(1)/(3) \pi (4)^(2)h \ or \  (1)/(3) \pi (4)^(2)(10) \ or \  (1)/(3) \pi (16)(10)

Multiply
(1)/(3) \pi(160) \ or \  (1)/(3) (3.14)(160) \ or \ 1.047(160)

Multiply
167.52 \ cm^(3)

This would make 167 \ cm^(3) the closest option.
The formula for the volume of a cone is π r²h/3. You are given the height and the diametre of the cone. You need the radius which is half the diametre (4). So, f you were to write this out, it would look like this:

3.14 × 4² × 10/3 

your answer is going to be 167 cm³

What is the equation line, in point-slope form, that passes through the points (-3,-1) And (-6,8)

Answers

Answer:

The equation line is y + 3x = -10.

Step-by-step explanation:

In this question, first we have to find the slope of the equation:

The formula which we use to find the slope is:

slope = m = (y2-y1)/(x2-x1)

m = (8 + 1) / (-6 + 3)

m = 9 / (-3)

m = -3.

Now, put the value of slope and the first points (-3, -1). You can use any of the given points:

y - y1 = m(x - x1)

y + 1 = -3(x + 3)

y + 1 = -3x - 9

y + 3x = -10

This is the eqution line.

You can check if this line is true or not by putting the other points in this equation which will give the same points like,

y + 3(-6) = -10

y - 18 = -10

y = 8

So, we get the same old points (-6, 8) which we have given.