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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P =$3000, r = 28% , t = 5 years
<span>A = 3000.e^(0.28)(5) = 3000.(4.0551) = $12,165.60</span>
Answer:
64 crates
Step-by-step explanation:
Smaller Cube Side Length = 2 1/2 feet, or, 2.5 feet
Larger Container (Cube) Side Length = 10 feet
We find volume of larger container and find volume of small crates. We divide the large volume by volume of each crate. This will give us number of crates we can fit.
Volume of Cube = x^3
Where x is the side length of the cube
Now,
Small Crate Volume = (2.5)^3 = 15.625 cubic feet
Large Container Volume = 10^3 = 1000 cubic feet
Number of crates that would fit = 1000/15.625 = 64
So, 64 crates will fit in the largest shipping container
Answer:
The answer would be 15. Your welcome my good sir
Step-by-step explanation:
<span>product of (3.7 × 104) and 2
</span>
answer is A.) 7.4 × 104