1. 3x - 2y = 0 => y = 3/2x
4x + 2y = 14<=> 4x + 2*(3/2x) = 14<=> 7x = 14<=> x = 2 & y = 3
2.3p + q = 7=> q = 7 - 3p
2p - 2q = -6<=> 2p - 2*(7-3p) = -6<=> 2p - 14 + 6p = -6<=> 8p = -6 + 14 = 8<=> p = 1 & q = 4
3.3x - 2y = 1=> x = (1+2y)/3
8x + 3y = 2<=> 8*(1+2y)/3 + 3y = 2<=> 8*(1+2y)/3 - 2 = -3y<=> 3*(8*(1+2y)/3 - 2) = -3*(3y)<=> 8*(1+2y) - 6 = -9y<=> 8 + 16y - 6 = -9y<=> 2 = -25y<=> y = -2/25 & x = 7/25
Answer:
Step-by-step explanation:
13. Find the total number of coinsss
8+4+3+1
=16
First pick:
1/16
Second pick:
1/15
So I guess your chances are 1/16+1/15 ?
=31/240
hopefully
14.
add the sockss
3+4+3
= 10
First pick
1/10
Second Pick
1/9
Add them:
19/90
Forgive me if this is wrong
:,)
Answer:
a^2
−10a+21
Step-by-step explanation:
(a−7)(a−3)
Apply the distributive property by multiplying each term of a−7 by each term of a−3.
a^2
−3a−7a+21
Combine −3a and −7a to get −10a.
a^2
−10a+21
2(2x + 1)
= 2 * 2x + 2
= 4x + 2
So, in conclusion, the answer to this question is a) 2 * 2x + 2
Answer:
88
Step-by-step explanation:
If there are 111 balls of 4 colors, if 100 are taken, and each at least 1 of each color appears, there has to be at least 12 of each color (if you take 100 marbles of 3 colors, all 11 of 1 color could still be in the box, so there has to be at least 12 of each). If there are 12 marbles of 3 colors, there would be 111-(12×3)=111-36-75 marbles of the last color. To get at least 1 marble in 3 colors add 75+12+1=88. If there are 12 marbles of 2 colors, there would be a combined total of 87 marbles for the other 2 colors. 87+1=88. If there is 12 marbles of 1 color, there would be a combined total of 99 marbles for the other 3 colors. If there was a divide of 97+1+1, the 2 colors with only 1 marble could have still been left in the box when 100 balls were taken. So there has to be at least 12 of the colors. So we get 111-12-12-12=75, which is the maximum number of marbles that 1 color can have. 75+12+1=88, our previous answer.
Hope this helps! (P.S. Please give me branliest, it will help me achieve my next rank.)