You can make some algebraic equations and solve it.
The first would be:

The second would be

You can then rearrange the second into

And subsitute it into the first like so:

After that, distribute the y into the parantheses.

Subtract the 21 on both sides and multiply by -1 on both sides:

You then can factor it into:

With Zero Product Property, we can determine y to be either -3 and 7. Since the variables are interchangable, you can say the same about x, just that whatever x is, y must be the other value.
Thus, the answer is 7 and -3.
Answer:
y = -
x + 
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Rearrange 4y = 5x + 3 into this form by dividing the 3 terms by 4
y =
x +
← in slope- intercept form
with slope m = 
Given a line with slope m then the slope of a line perpendicular to it is
= -
= -
= -
, thus
y = -
x + c ← is the partial equation
To find c substitute (3, - 2) into the partial equation
- 2 = -
+ c ⇒ c = - 2 +
= 
y = -
x +
← equation of perpendicular line
Answer:
10
Step-by-step explanation:
6 is the width not the length
Answer:
they re 9 members dat came
Answer:
Part A: impossible
Part B: Either equal or blue
Part C: 9 green and 2 blue were added
Step-by-step explanation:
Part A:
The only colors included in this problem are red, blue, and green. There is no black colored pencil, therefore, it is impossible to get one from the box.
Part B:
I'm not sure what you're asking in this question, but I will give you the two choices. If it is before the additional 11 colored pencils are added to the box, the chance of drawing a red and the chance of drawing a blue will be equal, because both of them have 11 of each color. If it is after the additional 11 colored pencils are added to the box, then the chance of drawing a blue colored pencil will be greater than the chance of drawing a red colored pencil. After the 11 colored pencils are added, there are 13 blue and 11 red. The blue is greater.
Part C:
The least number of green colored pencils added has to be 9, because the chance of drawing a green pencil is now greater than the chance of drawing a red pencil. If we add 8 more green pencils, the likelihood would be the same. Therefore, the number of green colored pencils added has to be at least 9. If we have the last 2 colored pencils be blue, then there would be 11 red, 13 blue, and 12 green. This fits all the conditions, therefore, adding 9 green colored pencils and 2 blue colored pencils is the answer.
I hope this helps and please mark me as brainliest!