-4(2y)+11y=15, -8y+11y=3y=15: y=5
X=2(5)=10
To check your work -4(10)+11(5)=15
15=15
Answer:
well can I see the choices?
We are given with
a1 = 2
r = 4
These are components of a geometric series. The first term is 2 and the common ratio is 4. To get the first six terms, we use the formula:
an = a1 r^(n-1)
a1 = 2 (4)^(1-1) = 2
a2 = 2 (4)^(2-1) = 8
a3 = 2 (4)^(3-1) = 32
a4 = 2 (4)^(4-1) = 128
a5 = 2 (4)^(5-1) = 512
a6 = 2 (4)^(6-1) = 2048
One possible system is
1x + 3y = 4
2x + 6y = 8
Note how 2 is twice as large as 1, 6 is twice as large as 3, and 8 is twice as large as 4.
In other words, the second equation is the result of multiplying both sides of the first equation by 2.
1x+3y = 4
2*(1x+3y) = 2*4
2x+6y = 8
Effectively the two equations in bold are the same which produces the same line. The two lines overlap perfectly to intersect infinitely many times. An intersection is a solution.
Answer:
20 green pencils
Step-by-step explanation:
The difference between black and green "ratio units" is 7 -5 = 2. The difference between red and blue ratio units is 3 -2 = 1. The sum of those differences is 2 +1 = 3 ratio units, which represent 12 pencils. Then each ratio unit represents 12/3 = 4 pencils.
The 5 ratio units of green pencils represent 5·4 pencils = 20 green pencils.
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If you need an equation, you can use k as the constant of proportionality. Then the sum of differences described is ...
(7k -5k) + (3k -2k) = 12
3k = 12 . . . . simplify
k = 4 . . . . . . divide by 3
5k = 5·4 = 20 . . . . . number of green pencils
