The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
<h3>How to determine the number of real zeros?</h3>
The equation of the function is given as:
Expand the function
Reorder the terms
Factor the expression
Factor out x -1
Expand
Factorize
](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Bx%28x%20%2B%203%29%20%2B%202%28x%20%2B%203%29%5D%28x%20-%201%29)
Factor out x + 2
The function has been completely factored and it has 3 linear factors
Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
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Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.
<u>Step-by-step explanation:</u>
As given by the statement in the problem,
Q may be the middle point, which cut the diagonal PA into 2 equal halves.
In rhombus, diagonals meet at right angles.
which means that PQ = QA
x+2 = 3x - 14
Grouping the terms, we will get,
3x -x = 14+ 2
2x = 16
dividing by 2 on both sides, we will get,
x = 16/2 = 8
8+2 = 3(8) - 14 = 10 = PQ or QA
8*8=64 well 2*2=4 hope it helps
Assume that pi is approx. 3.14. Then 86.92 = approx. 27 2/3 times pi, or
86.92 is approx. equal to 83pi/3.
Answer:
-2/45
Step-by-step explanation:
-3/5 - (-5/9)
Subtracting a negative is like adding
-3/5 + 5/9
Get a common denominator of 45
-3/5*9/9 + 5/9 *5/5
-27/45 + 25/45
-2/45
