Answer:
a) 
b) 
c) 
Step-by-step explanation:
<u>For the question a *</u> you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.
This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.
Then write the polynomial in factored form.
Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.
<u>For question b, do the same procedure</u>.
Degree: 3
Zeros: -5/2, 4/5, 6.
The factors are
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<u>Finally for the question c we have</u>
Degree: 5
Zeros: -3, 1, 4, -1
Multiplicity 2 in -1
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Answer:
1
Step-by-step explanation:
The common factors are:
1
The Greatest Common Factor:
GCF = 1
Solution
Teh factors of 100 are:
1, 2, 4, 5, 10, 20, 25, 50, 100
Teh factors of 139 are:
1, 139
The common factors are:
1
The Greatest Common Factor:
GCF = 1
I got A, -32, but it has been awhile since I have learned this in my math class. You might want to wait for someone else to verify this answer.
For this problem, all you need to do is find the three #'s that add up to 156.
So, lets look at the answers and add them up.
A. 50, 52, 54
50 + 52 + 54 = 156
B. 51,52,53
51 + 52 + 53 = 156
C. 49,50,51
49 + 50 + 51 = 150
D. 49,51,53
49 + 51 + 53 = 153
We get the answers (50,52,54) and (51,52,53)
Now, consecutive numbers are numbers that in order, like 1,2,3.
Therefore, the answer is (51,52,53)
T / 3/4 ; t = 9 3/4
9 3/4 / 3/4
9 3/4 = 39/4
39/4 / 3/4
39/4 x 4/3
39 x 4 = 156
4 x 3 = 12
156/12 / 4/4 = 39/3 = 13
