A
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There are 3 possibilities for a marble (3 different marbles) and 2 possibilities for a coin (heads or tails). We multiply these to get (3)(2) = 6 outcomes.
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Hope this helps!
==jding713==
Answer:
a = 2, b = - 3, c = - 8
Step-by-step explanation:
Expand f(x) = a(x + b)² + c and compare coefficients of like terms, that is
a(x + b)² + c ← expand (x + b)² using FOIL
= a(x² + 2bx + b²) + c ← distribute parenthesis by a
= ax² + 2abx + ab² + c
Compare like terms with f(x) = 2x² - 12x + 10
Compare coefficients x² term
a = 2
Compare coefficients of x- term
2ab = - 12, substitute a = 2
2(2)b = - 12
4b = - 12 ( divide both sides by 4 )
b = - 3
Compare constant term
ab² + c = 10 , substitute a = 2, b = - 3
2(- 3)² + c = 10
18 + c = 10 ( subtract 18 from both sides )
c = - 8
Then a = 2, b = - 3, c = - 8
Answer:
6+(-3) is 3 units from 6 in the right direction
Step-by-step explanation:
Answer:
A) The value of a is <u>29</u>.
B) The value of b is <u>greater than 29</u>.
C) In both part A and part B we have used a common property which is addition property and that we have add 9 on both side of equation in both parts.
D) The value of a in part A is equal to 29 whereas in part B the value of b is greater than 29.
Step-by-step explanation:
Solving for Part A.
Given,
We have to solve for a.
By using addition property of equality, we will add both side by 9;
Hence the value of a is <u>29</u>.
Solving for Part B.
Given,
We have to solve for b.
By using addition property of inequality, we will add both side by 9;
Hence the value of b is <u>greater than 29</u>.
Solving for Part C.
In both part A and part B we have used a common property which is addition property and that we have add 9 on both side of equation in both parts.
Solving for Part D.
The value of a in part A is equal to 29 whereas in part B the value of b is greater than 29.