A + b = 45
a - b = 21
a = 21 + b
Use substitution
21 + b + b = 45
2b = 24
b = 12
45 - 12 = 33
The numbers are 33 and 12
Answer:
y = 3/4 or y = -3/5
Step-by-step explanation:
Solve for y:
(8 y - 6) (10 y + 6) = 0
Hint: | Find the roots of each term in the product separately.
Split into two equations:
8 y - 6 = 0 or 10 y + 6 = 0
Hint: | Look at the first equation: Factor the left hand side.
Factor constant terms from the left hand side:
2 (4 y - 3) = 0 or 10 y + 6 = 0
Hint: | Divide both sides by a constant to simplify the equation.
Divide both sides by 2:
4 y - 3 = 0 or 10 y + 6 = 0
Hint: | Isolate terms with y to the left hand side.
Add 3 to both sides:
4 y = 3 or 10 y + 6 = 0
Hint: | Solve for y.
Divide both sides by 4:
y = 3/4 or 10 y + 6 = 0
Hint: | Look at the second equation: Factor the left hand side.
Factor constant terms from the left hand side:
y = 3/4 or 2 (5 y + 3) = 0
Hint: | Divide both sides by a constant to simplify the equation.
Divide both sides by 2:
y = 3/4 or 5 y + 3 = 0
Hint: | Isolate terms with y to the left hand side.
Subtract 3 from both sides:
y = 3/4 or 5 y = -3
Hint: | Solve for y.
Divide both sides by 5:
Answer: y = 3/4 or y = -3/5
Answer:
C. y<3
Step-by-step explanation:
We are required to determine the solution set of the inequality

Step 1: Add 9 to both sides

Step 2: Divide both sides by 15

The solution set of the given inequality is: y<3.
The correct option is C.
Answer:
Probability that a student chosen randomly from the class plays basketball or baseball is
or 0.76
Step-by-step explanation:
Given:
Total number of students in the class = 30
Number of students who plays basket ball = 19
Number of students who plays base ball = 12
Number of students who plays base both the games = 8
To find:
Probability that a student chosen randomly from the class plays basketball or baseball=?
Solution:
---------------(1)
where
P(A) = Probability of choosing a student playing basket ball
P(B) = Probability of choosing a student playing base ball
P(A \cap B) = Probability of choosing a student playing both the games
<u>Finding P(A)</u>
P(A) = 
P(A) =
--------------------------(2)
<u>Finding P(B)</u>
P(B) = 
P(B) =
---------------------------(3)
<u>Finding
</u>
P(A) = 
P(A) =
-----------------------------(4)
Now substituting (2), (3) , (4) in (1), we get


