This question above is incomplete
Complete Question
Jim had 103 red and blue marbles. After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles. How many blue marbles did he have originally?
Answer:
70 Blue marbles
Step-by-step explanation:
Let red marbles = R
Blue marbles = B
Step 1
Jim had 103 red and blue marbles.
R + B = 103.......Equation 1
R = 103 - B
Step 2
After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles
2/5 of B to Samantha
Jim has = B - 2/5B = 3/5B left
He also gave 15 red marbles to Samantha
= R - 15
The ratio of what Jim has left
= Red: Blue
= 3:7
= 3/7
Hence,
R - 15/(3/5)B = 3/7
Cross Multiply
7(R - 15) = 3(3/5B)
7R - 105 = 3(3B/5)
7R - 105 = 9B/5
Cross Multiply
5(7R - 105) = 9B
35R - 525 = 9B............ Equation 2
From Equation 1, we substitute 103 - B for R in Equation 2
35(103 - B) - 525 = 9B
3605 - 35B - 525 = 9B
Collect like terms
3605 - 525 = 9B + 35B
3080 = 44B
B = 3080/44
B = 70
Therefore, Jim originally had 70 Blue marbles.
C I think I hope this is correct
Hey there !
The answer would be A. n = t + 2
The sequence is adding two to the last term in the sequence, or t.
Hope this helps !
Answer:31/40
Step-by-step explanation:
To add you have to have a common denominator. To get that you can multiply each fraction by one in different ways. 2/5 *8/8= 16/40 and 3/8* 5/5= 15/40 and then add those two to give you 31/40.
