The setup boxes in the synthetic division are (b)
<h3>How to determine the setup boxes?</h3>
The dividend is given as:
x^3 + 4x^2 + x - 6
The divisor is given as:
x - 2
Set the divisor to 0
x - 2 = 0
Solve for x
x = 2
Remove the variables in the dividend
1 + 4 + 1 - 6
Remove the arithmetic signs
1 4 1 - 6
So, the setup is:
2 | 1 4 1 - 6
Hence, the setup boxes are (b)
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<span>If Bruce does not have beans for supper, then it is not Friday.</span>
3/5*3
pretend that 3 has a denominator which is 1
3/5*3/1
mutiply the numerators together
3*3= 9
mutiply the denominators together
5*1= 5
Answer:
9/5, 1.8 and 1 4/5
Make the fractions improper:
33/4 + 14/5
Find common denominator:
165/20 + 56/20
Add:
221/20
Answer:
11 1/20
