So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:
Next, square both sides:
Next, subtract x and add 2 to both sides of the equation:
Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:
Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:
Now you can rewrite the equation as 
Now, apply the Zero Product Property and solve for x as such:
Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:
Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>
The answer is false due to the object is changes shape not size
Answer:
Option b
Step-by-step explanation:
To represent this proportion using fractions, let the numerators be the number of waffles sold and denominator, the number of sugar cones.
Thus, the proportion that can be used to represent the cone sales is
14/6 = 7/3
Answer:
The first one is 0, the second one is ARN (All real numbers)
Step-by-step explanation:
1. Anything raised to the "zeroth" power is always 1
2. Because 7 is being raised to the zeroth power, the number inside the parentheses becomes 1. Now x can equal anything, 1 times itself will always equal 1, no matter how many times you multiply them.
Answer:
<h2>4/3 Joules </h2>
Step-by-step explanation:
Work is said to be done when force applied to an object causes the object to move through a distance.
Work done = Force * perpendicular distance.
Given Force F = xy i + (y-x) j and a straight line (-1, -2) to (1, 2)
First we need to get the equation of the straight line given.
Given the slope intercept form y = mx+c
m is the slope
c is the intercept
m = y₂-y₁/x₂-x₁
m = 2-(-2)/1-(-1)
m = 4/2
m = 2
To get the slope we will substtutte any f the point and the slope into the formula y = mx+c
Using the point (1,2)
2 = 2+c
c = 0
y = 2x
Substituting y = 2x into the value of the force F = xy i + (y-x) j we will have;
F = x(2x) i + (2x - x) j
Using the coordinate (1, 2) as the value of s
![W = \int\limits^a_b ({2x^2 i + x j}) \, (i+2j)\\W = \int\limits^a_b ({2x^{2}+2x }) \, dx \\W = [\frac{2x^{3} }{3} +x^{2} ]\left \ x_2=1} \atop {x_1=-1}} \right.\\W = (2(1)^3/3 + 1^2) - (2(-1)^3/3 + (-1)^2)\\W =(2/3+1) - (-2/3+1)\\W = 2/3+2/3+1-1\\W = 4/3 Joules](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5Ea_b%20%28%7B2x%5E2%20i%20%2B%20x%20j%7D%29%20%5C%2C%20%28i%2B2j%29%5C%5CW%20%3D%20%5Cint%5Climits%5Ea_b%20%28%7B2x%5E%7B2%7D%2B2x%20%7D%29%20%5C%2C%20dx%20%5C%5CW%20%3D%20%5B%5Cfrac%7B2x%5E%7B3%7D%20%7D%7B3%7D%20%2Bx%5E%7B2%7D%20%5D%5Cleft%20%5C%20x_2%3D1%7D%20%5Catop%20%7Bx_1%3D-1%7D%7D%20%5Cright.%5C%5CW%20%3D%20%282%281%29%5E3%2F3%20%2B%201%5E2%29%20-%20%20%282%28-1%29%5E3%2F3%20%2B%20%28-1%29%5E2%29%5C%5CW%20%3D%282%2F3%2B1%29%20-%20%28-2%2F3%2B1%29%5C%5CW%20%3D%202%2F3%2B2%2F3%2B1-1%5C%5CW%20%3D%204%2F3%20Joules)
