Answer:
C
Step-by-step explanation:
write an equation for the costs:
if x is the number of sodas
and y is the number of waters
2.75x + 2y <= 15
(<= is less than or equal to)
if we substitute 3 for y
we get 2.75x + 2(3) <= 15
2.75x + 6 <= 15
2.75x <= 9
9 / 2.75 = 3.2727
however, you cannot buy part of a soda
so, round to 3
you also cannot buy negative sodas
so, the answer is C
Answer: complex equations has n number of solutions, been n the equation degree. In this case:
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi11%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi101%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi191%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi281%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi78%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi168%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi258%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi348%2C75%C2%B0%7D)
Step-by-step explanation:
I start with a variable substitution:
![Z^{4} = X](https://tex.z-dn.net/?f=Z%5E%7B4%7D%20%3D%20X)
Then:
![2X^{2}-2X+1=0](https://tex.z-dn.net/?f=2X%5E%7B2%7D-2X%2B1%3D0)
Solving the quadratic equation:
![X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}](https://tex.z-dn.net/?f=X_%7B1%7D%20%3D%5Cfrac%7B2%2B%5Csqrt%7B4-4%2A2%2A1%7D%20%7D%7B2%2A2%7D%20%5C%5CX_%7B2%7D%20%3D%5Cfrac%7B2-%5Csqrt%7B4-4%2A2%2A1%7D%20%7D%7B2%2A2%7D)
![X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.](https://tex.z-dn.net/?f=X%3D%5Cleft%20%5C%7B%20%7B%7B0%2C5%2B0%2C5i%7D%20%5Catop%20%7B0%2C5-0%2C5i%7D%7D%20%5Cright.)
Replacing for the original variable:
![Z=\sqrt[4]{0,5+0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5%2B0%2C5i%7D)
or ![Z=\sqrt[4]{0,5-0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5-0%2C5i%7D)
Remembering that complex numbers can be written as:
![Z=a+ib=|Z|e^{ic}](https://tex.z-dn.net/?f=Z%3Da%2Bib%3D%7CZ%7Ce%5E%7Bic%7D)
Using this:
![Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.](https://tex.z-dn.net/?f=Z%3D%5Cleft%20%5C%7B%20%7B%7B%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi45%C2%B0%7D%20%7D%20%5Catop%20%7B%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi-45%C2%B0%7D%20%7D%7D%20%5Cright.)
Solving for the modulus and the angle:
![Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.](https://tex.z-dn.net/?f=Z%3D%5Cleft%20%5C%7B%20%7B%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi45%7D%7D%20%7D%20%5Catop%20%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi-45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi-45%7D%7D%20%7D%7D%20%5Cright.)
The possible angle respond to:
![RAng_{12...n} =\frac{Ang +360*(i-1)}{n}](https://tex.z-dn.net/?f=RAng_%7B12...n%7D%20%3D%5Cfrac%7BAng%20%2B360%2A%28i-1%29%7D%7Bn%7D)
Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"
In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º
Obtaining 8 different angles, therefore 8 different solutions.
Answer:
25
Step-by-step explanation:
<u>Formula</u>
![B = 25(\frac{M}{10} +\frac{3}{4} )](https://tex.z-dn.net/?f=B%20%3D%2025%28%5Cfrac%7BM%7D%7B10%7D%20%2B%5Cfrac%7B3%7D%7B4%7D%20%29)
Where M meters of granola makes B granola bars.
<u>Given:</u>
M = 2.5 meters
Now substitute that value in the formula and calculate.
![B=25(\frac{2.5}{10} +\frac{3}{4} )= 25(1) = 25](https://tex.z-dn.net/?f=B%3D25%28%5Cfrac%7B2.5%7D%7B10%7D%20%2B%5Cfrac%7B3%7D%7B4%7D%20%29%3D%2025%281%29%20%3D%2025)
Therefore, 2.5 meters of granola makes 25 granola bars.
B = 25
Answer:
y = -4x² + 32x - 48
Step-by-step explanation:
The standard form of a quadratic equation is
y = ax² + bx + c
We must find the equation that passes through the points:
(2, 0), (6,0), and (3, 12)
We can substitute these values and get three equations in three unknowns.
0 = a(2²) + b(2) + c
0 = a(6²) + b(6) + c
12 = a(3²) + b(3) + c
We can simplify these to get the system of equations:
(1) 0 = 4a + 2b + c
(2) 0 = 36a + 6b + c
(3) 12 = 9a + 3b + c
Eliminate c from equations (1) and (2). Subtract (1) from (2).
(4) 0 = 32a + 4b
Eliminate c from equations (2) and (3). Subtract (3) from (2).
(5) -12 = 27a - 3b
Simplify equations (4) and (5).
(6) 0 = 8a + b
(7) -4 = 9a - b
Eliminate b by adding equations (6) and (7).
(8) a = -4
Substitute (4) into (6).
0 = -32 + b
(9) b = 32
Substitute a and b into (1)
0 = 4(-4) + 2(32) + c
0 = -16 + 64 + c
0 = 48 + c
c = -48
The coefficients are
a= -4, b = 32, c = -48
The quadratic equation is
y = -4x² + 32x - 48
The diagram below shows the graph of your quadratic equation and the three points through which it passes.