Some rational numbers are not integers. Is The False Option....
Answer:
We have: g(x)=(x+2)(x-1)(x-2)
multiply the 3 terms:
- (x+2)(x-1)(x-2)
- (x²-4)(x-1)
- x^3 -x²-4x+4
This is a polynimial function with 2 vertices
The roots are approximatively:
0.8 and -1.5
their images are:
1.9 and 8.8
plot these two vertices and draw some other points
here is a drawing:
Answer:
B) Angle 2 and Angle 3
Step-by-step explanation:
Angle 2 and Angle 3 are equal since they're vertical angles and aren't supplementary
Angle 7 and 8 are supplementary
Angle 1 and 7 are supplementary
Honestly, i have no clue with angle 4 and angle 4. I just know that vertical angles aren't supplementary because they're equal to each other so I'll assume this is the case as well.
Answer:
The length=16cm and the width=8cm.
Step-by-step explanation:
Given that the length is twice the breadth or width of the rectangle
Let's assume that the breadth of the rectangle is x.
Thus the length is 2x.
Given perimeter=48cm
The formula for the perimeter of a rectangle is 2(l+b) where l is length and b is breadth.
2(x+2x)=48
(3x)=48/2
3x=24
x=8cm
2x=16cm
Answer: ![\frac{\sqrt[4]{10xy^3}}{2y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D)
where y is positive.
The 2y in the denominator is not inside the fourth root
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Work Shown:
![\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%7D%7B8y%7D%7D%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%2A2y%5E3%7D%7B8y%2A2y%5E3%7D%7D%5C%20%5C%20%5Ctext%7B....%20multiply%20top%20and%20bottom%20by%20%7D%202y%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B10xy%5E3%7D%7B16y%5E4%7D%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B16y%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20break%20up%20the%20fourth%20root%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20rewrite%20%7D%2016y%5E4%20%5Ctext%7B%20as%20%7D%20%282y%29%5E4%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D%20%5C%20%5C%20%5Ctext%7B...%20where%20y%20is%20positive%7D%5C%5C%5C%5C%5C%5C)
The idea is to get something of the form
in the denominator. In this case, 
To be able to reach the
, your teacher gave the hint to multiply top and bottom by
For more examples, search out "rationalizing the denominator".
Keep in mind that
only works if y isn't negative.
If y could be negative, then we'd have to say
. The absolute value bars ensure the result is never negative.
Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.