Answer:
B. (2,8), because both equations intersect at this point
Answer: Insufficient Information
The probability of selecting a red ball, a blue ball or a white ball is unknown.
Step-by-step explanation:
The probability of selecting a red ball, a blue ball or a white ball is unknown.
There could be 13 red balls, 1 blue ball and 11 white balls in the bag. There could be 3 red balls, 2 blue balls and 20 white balls in the bag. It is impossible to say with the information given have even numbers on them. The same thing applies to the number of balls with numbers or even numbers on them. The question does not specify how many white balls have each number on them.
Step-by-step explanation:
let d+e=2 be equation 1
let d-e=4 be equation 2
lets make d the subject in equa 1
d = 2-e equation 3
put equation 3 in 2
2-e=4
e=2-4
e=-2
put e=-2 in equa 1
d-2=2
d=4
e is -2 and d is 4
It would be 7lbs, you divide the total amount by the amount per pound and you get the weight.
Answer:
1/5 + 2/5 i sqrt(6) = .2 + .98i
1/5 - 2/5 i sqrt(6) = .2 - .98i
Step-by-step explanation:
5z^2−9z=−7z−5
We need to get all the terms on one side (set the right side equal to zero)
Add 7z to each side
5z^2−9z+7z=−7z+7z−5
5z^2−2z=−5
Add 5 to each side
5z^2−2z+5=−5 +5
5z^2−2z+5=0
This is in the form
az^2 +bz+c = 0 so we can use the quadratic formula
where a = 5 b = -2 and c = 5
-b± sqrt(b^2-4ac)
-------------------------
2a
-(-2)± sqrt((-2)^2-4(5)5)
-------------------------
2(5)
2± sqrt(4-100)
-------------------------
10
2± sqrt(-96)
-------------------------
10
2± sqrt(16)sqrt(-1) sqrt(6)
-------------------------
10
2± 4i sqrt(6)
-------------------------
10
1/5 ± 2/5 i sqrt(6)
Splitting the ±
1/5 + 2/5 i sqrt(6) = .2 + .98i
1/5 - 2/5 i sqrt(6) = .2 - .98i
