90^2-54^2=5184
root 5184=72
pr=72
(2,-1)(8,4)
slope = (4-(-1) / (8-2) = (4 + 1)/6 = 5/6
y - y1 = m(x - x1)
(8,4)...x1 = 8 and y1 = 4
slope(m) = 5/6
now we sub
y - 4 = 5/6(x - 8) <==point-slope form
y - 4 = 5/6x - 20/3
y = 5/6x - 20/3 + 4
y = 5/6x -20/3 + 12/3
y = 5/6x - 8/3
-5/6x + y = - 8/3...multiply by -6
5x - 6y = 16 <==standard form
Answer:
The sum of the two remaining numbers, x and y = 60
Question:
The question isn't clear enough as some information have been omitted. Let's consider the following:
Model with Math. The average of six numbers is 18. If the average of four numbers is 12. What must be the sum of the two remaining numbers, x and y?
Write an equation to show how to find this sum.
Step-by-step explanation:
Mathematical models are applied to represent things in the real world in order to solve problems.
The formula we would use to solve this problem is an example of a mathematical model.
Types of mathematical model we can use include equations and graphs.
Using equations:
Average of six numbers = 18
Average of four of the numbers = 12
Total sum of the four numbers = 4×12 = 48
the two unknown numbers are x and y
Average of six numbers = (Sum of all 6 numbers)/6
=(Total sum of four numbers + x + y)/6
(48 + x + y)/6 = 18
The equation that shows how to find the sum:
(1/6)(48 + x + y) = 18
48 + x + y = 18×6
48 + x + y = 108
x + y = 108-48
x+y = 60
The sum of the two remaining numbers, x and y = 60
<h2><u>
PLEASE MARK BRAINLIEST!</u></h2>
Answer:
I don't know if this is right, but here goes nothing!
Step-by-step explanation:
It says (in small font) that the scale factor is 0.5. So that being said, what are the dilated coordinates of R?
If R is (1, 2), all we have to do is multiply both the x-value and the y-value by 0.5 to get our answer.
1 × 0.5 = 
2 × 0.5 = 1
So our answer is ( , 1).
<h3>The coordinates of R', or the dilated form of R, are (
, 1)</h3>
I hope this helps!
- sincerelynini
No 'real' numbers can do that job.
The two numbers are complex.
They are ...
<em>1 + i √3</em>
and
<em>1 - i √3</em> .
