5 3/4 + 1 3/4 = 23/4 + 7/4 = 30/4......total fruit juice
(30/4) / 1/2 =
30/4 * 2 =
60/4 =
15 <=== Abby can make 15 popsicles :)
For this case we have the following equation:
![\sqrt [4] {2x-8} + \sqrt [4] {2x + 8} = 0](https://tex.z-dn.net/?f=%5Csqrt%20%5B4%5D%20%7B2x-8%7D%20%2B%20%5Csqrt%20%5B4%5D%20%7B2x%20%2B%208%7D%20%3D%200)
If we subtract both sides of the equation ![\sqrt [4] {2x + 8}](https://tex.z-dn.net/?f=%5Csqrt%20%5B4%5D%20%7B2x%20%2B%208%7D)
we have:
![\sqrt [4] {2x-8} = - \sqrt [4] {2x + 8}](https://tex.z-dn.net/?f=%5Csqrt%20%5B4%5D%20%7B2x-8%7D%20%3D%20-%20%5Csqrt%20%5B4%5D%20%7B2x%20%2B%208%7D)
To eliminate the radical we raise both sides of the equation to the fourth power:
![(\sqrt [4] {2x-8}) ^ 4 = (- \sqrt [4] {2x + 8}) ^ 4](https://tex.z-dn.net/?f=%28%5Csqrt%20%5B4%5D%20%7B2x-8%7D%29%20%5E%204%20%3D%20%28-%20%5Csqrt%20%5B4%5D%20%7B2x%20%2B%208%7D%29%20%5E%204)
Answer:
Option D
Applying parallel line postulate twice
36degree=3x degree +2y degree
126 degrees=12 x degree + 2y degree
So, the two equations are:
3x + 2y = 36
and
12x + 2y = 126
Eliminate x to solve for y
-(3x + 2y)= -(36)
-3x - 2y = -36
Combine the equations
(-3x - 2y)+ (12x + 2y) = (-36) + (126)
-3x -2y + 12x + 2y = -36 + 126
9x = 90
x = 10
Plug value of x back into to the equations to solve for y
3(10) + 2y = 36
30 + 2y = 36
2y =6
y = 3
Optional
Plug x into the other equation to check for error
12(10) +2y = 126
120 + 2y =126
2y = 6
y = 3
Answer:
B. (-2,3)
Step-by-step explanation:
-3x+5y = 21
x=2y-8
Substitution
-3(2y-8)+5y =21
-6y+24+5y=21
-y+24=21
-y=-3
y=3
Plug-in other equation
2*3-8=x
6-8=x
x=-2
<h2>f</h2><h2>(</h2><h2>x</h2><h2>)</h2><h2>=</h2><h2>x</h2><h2>3</h2><h2>−</h2><h2>9</h2><h2>x</h2><h2>2</h2><h2>+</h2><h2>24</h2><h2>x</h2><h2>−</h2><h2>10</h2><h2>Taking first derivative of</h2><h2> </h2><h2> </h2><h2>f</h2><h2>(</h2><h2>x</h2><h2>)</h2><h2>f</h2><h2>′</h2><h2>(</h2><h2>x</h2><h2>)</h2><h2>=</h2><h2>3</h2><h2>x</h2><h2>2</h2><h2>−</h2><h2>18</h2><h2>x</h2><h2>+</h2><h2>24</h2><h2>For finding critical points substituting</h2><h2> </h2><h2>f</h2><h2>′</h2><h2>(</h2><h2>x</h2><h2>)</h2><h2>=</h2><h2>0</h2><h2>f</h2><h2>′</h2><h2>(</h2><h2>x</h2><h2>)</h2><h2>=</h2><h2>0</h2><h2> </h2><h2>⇒</h2><h2>3</h2><h2>x</h2><h2>2</h2><h2>−</h2><h2>18</h2><h2>x</h2><h2>+</h2><h2>24</h2><h2>=</h2><h2>0</h2><h2> </h2><h2>⇒</h2><h2>3</h2><h2>(</h2><h2>x</h2><h2>2</h2><h2>−</h2><h2>6</h2><h2>x</h2><h2>+</h2><h2>8</h2><h2>)</h2><h2>=</h2><h2>0</h2><h2>(</h2><h2>x</h2><h2>−</h2><h2>4</h2><h2>)</h2><h2>(</h2><h2>x</h2><h2>−</h2><h2>2</h2><h2>)</h2><h2>=</h2><h2>0</h2><h2>After solving the value of x is</h2><h2>x</h2><h2>=</h2><h2>2</h2><h2>,</h2><h2>4</h2><h2>Thus critical points at</h2><h2> </h2><h2> </h2><h2>x</h2><h2>=</h2><h2>2</h2><h2>,</h2><h2>4</h2>
