How do we graph anything? Make a table of values for x and y and then plot each point. After plotting each point on the xy-plane, connect each point with a straight line or curve (depending on the function).
In this case, we must first isolate y.
y = (-4/3)x + 8y
y - 8y = (-4/3)x
-7y = (-4/3)x
y = (-4/3)x ÷ (-7)
y = (4/21)x
Now follow the steps above.
Answer:




Step-by-step explanation:
Given
See attachment for complete question
Required
Match equivalent expressions
Solving (a):

The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:

Solving (b):

The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:

Solving (c):

The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:

Solving (d):

The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:

Answer:
whatever% of anything, is just (whatever/100) * anything.
there were a total of 560000 votes, now, off those some are valid and some are invalid.
we know 15% of that are invalid, that simply means that 85% are valid, since 85% + 15% is the whole thing, or 100%, if 15% are not good, the other 85% are the good ones.
how many are the valid ones anyway? well, 85% of 560000, which is just (85/100) * 560000.
we know that off those valid ones, the candidate got 75% of those, so how much is 75% of that?
well, (75/100) * [ (85/100) * 560000 ].
Step-by-step explanation:
No. he used 50% on the rides, and we have to convert the fraction to a percent by dividing 4&1 which equals 0.25, then we multiply that by 100 to get the percentage he used on video games which equals 25%. 50+25= 75% used tickets for both rides and video games. So then we subtract 75 from 100 and get 25% which means he used 25% of the tickets on batting cages. Your answer is no, he used 25% of his tickets on the batting cages.