The possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are ![\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cpm%5C%7B1%2C2%2C4%2C%5Cfrac%2013%2C%20%5Cfrac%2023%2C%5Cfrac%7B4%7D%7B3%7D%7D%5C%7D)
<h3>How to determine the possible zeros?</h3>
The function is given as:
f(x) = 3x^6 + 4x^3 -2x^2 + 4
The leading coefficient of the function is:
p = 3
The constant term is
q = 4
Take the factors of the above terms
p = 1 and 3
q = 1, 2 and 4
The possible zeros are then calculated as:
![\mathbf{Zeros = \pm\frac{Factors\ of\ q}{Factors\ of\ p}}](https://tex.z-dn.net/?f=%5Cmathbf%7BZeros%20%3D%20%5Cpm%5Cfrac%7BFactors%5C%20of%5C%20q%7D%7BFactors%5C%20of%5C%20p%7D%7D)
So, we have:
![\mathbf{Zeros = \pm\frac{1,2,4}{1,3}}](https://tex.z-dn.net/?f=%5Cmathbf%7BZeros%20%3D%20%5Cpm%5Cfrac%7B1%2C2%2C4%7D%7B1%2C3%7D%7D)
Expand
![\mathbf{Zeros = \pm\frac{1,2,4}{1},\pm\frac{1,2,4}{3}}](https://tex.z-dn.net/?f=%5Cmathbf%7BZeros%20%3D%20%5Cpm%5Cfrac%7B1%2C2%2C4%7D%7B1%7D%2C%5Cpm%5Cfrac%7B1%2C2%2C4%7D%7B3%7D%7D)
Solve
![\mathbf{Zeros = \pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}](https://tex.z-dn.net/?f=%5Cmathbf%7BZeros%20%3D%20%5Cpm%5C%7B1%2C2%2C4%2C%5Cfrac%2013%2C%20%5Cfrac%2023%2C%5Cfrac%7B4%7D%7B3%7D%7D%5C%7D)
Hence, the possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are ![\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cpm%5C%7B1%2C2%2C4%2C%5Cfrac%2013%2C%20%5Cfrac%2023%2C%5Cfrac%7B4%7D%7B3%7D%7D%5C%7D)
Read more about rational root theorem at:
brainly.com/question/9353378
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Answer:
D. ![\frac{{7}^{11} }{ {4}^{11} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%7B7%7D%5E%7B11%7D%20%7D%7B%20%7B4%7D%5E%7B11%7D%20%7D%20)
Step-by-step explanation:
![( \frac{7}{4}) {}^{11} = \frac{7 {}^{11} }{4 {}^{11} }](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B7%7D%7B4%7D%29%20%7B%7D%5E%7B11%7D%20%3D%20%5Cfrac%7B7%20%7B%7D%5E%7B11%7D%20%7D%7B4%20%7B%7D%5E%7B11%7D%20%7D)
Since we are multiplying an exponential number to a fraction the answer is going to be
![\frac{7 {}^{11} }{4 {}^{11} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B7%20%7B%7D%5E%7B11%7D%20%7D%7B4%20%7B%7D%5E%7B11%7D%20%7D%20)
Hope this helps ❤❤❤ ;)
Answer:
592,000
Step-by-step explanation:
The new dimensions are 500, 280, and 200. Multiply 2(500*280 + 500*200 + 280*200) to get the answer
The distance of plane from the starting point is 69.46 km.
Given that the plane first flies 35 km in the direction N30W and then 60 km in the direction N60E.
We have to find the distance of plane from starting point.
We have to use pythagoras theorem in this which says that the square of hypotenuse is equal to sum of squares of the base and perpendicular.
![H^{2} =P^{2} +B^{2}](https://tex.z-dn.net/?f=H%5E%7B2%7D%20%3DP%5E%7B2%7D%20%2BB%5E%7B2%7D)
From the figure we can say that P=35 km .and B=60 km
In this we have to find the value of H when P=35 and B=60
![H^{2} =35^{2} +60^{2}](https://tex.z-dn.net/?f=H%5E%7B2%7D%20%3D35%5E%7B2%7D%20%2B60%5E%7B2%7D)
=1225+3600
=6946
H=69.46
Hence the distance of plane from starting point is 69.46 km.
Learn more about pythagoras theorem at brainly.com/question/343682
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800 is 100% and you want to increase by 40%
Add together
100% + 40% = 140%
140% expressed as a decimal
140/100 = 1.4
Multiply 800 by the multiplier
800 x 1.4 = 1120