Answer:
The equation that can be used to find the width is 2(13 + w) = 42
The width of the rectangle is 8 inches
Step-by-step explanation:
The formula of the perimeter of a rectangle is P = 2(l + w), where l is its length and w is its width
∵ One rectangle has a length of 13 inches
∴ l = 13 inches
∵ Its perimeter is 42 inches
∴ P = 42 inches
- Substitute the values of l and P in the formula of the perimeter
∵ 42 = 2(13 + w)
The equation that can be used to find the width is 2(13 + w) = 42
To find W divide both sides by 2
∴ 21 = 13 + w
- Subtract 13 from both sides
∴ 8 = w
The width of the rectangle is 8 inches
Answer:
12
Step-by-step explanation:
Remark
I can't use a symbol for the currency. I'll just get the answer in numbers. You can add the symbol.
Solution
First day: 50
Second day: - 20
Third day: - 18
50 - 20 - 18 = 12
Answer:
1/2
Step-by-step explanation:
(1/16)^x
Let x = 1/4
(1/16)^ 1/4
Rewriting 16 as 2^4
(1/2^4)^ 1/4
We know that 1 / a^b = a^-b
(2 ^ -4)^ 1/4
We know that a^b^c = a^(b*c)
2^(-4*1/4)
2^-1
We know that a^-b = 1/ a^b
2^-1 = 1/2^1 = 1/2
We have to calculate the fourth roots of this complex number:
![z=9+9\sqrt[]{3}i](https://tex.z-dn.net/?f=z%3D9%2B9%5Csqrt%5B%5D%7B3%7Di)
We start by writing this number in exponential form:
![\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20r%3D%5Csqrt%5B%5D%7B9%5E2%2B%289%5Csqrt%5B%5D%7B3%7D%29%5E2%7D%20%5C%5C%20r%3D%5Csqrt%5B%5D%7B81%2B81%5Ccdot3%7D%20%5C%5C%20r%3D%5Csqrt%5B%5D%7B81%2B243%7D%20%5C%5C%20r%3D%5Csqrt%5B%5D%7B324%7D%20%5C%5C%20r%3D18%20%5Cend%7Bgathered%7D)
![\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Carctan%20%28%5Cfrac%7B9%5Csqrt%5B%5D%7B3%7D%7D%7B9%7D%29%3D%5Carctan%20%28%5Csqrt%5B%5D%7B3%7D%29%3D%5Cfrac%7B%5Cpi%7D%7B3%7D)
Then, the exponential form is:

The formula for the roots of a complex number can be written (in polar form) as:

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.
To simplify the calculations, we start by calculating the fourth root of r:
![r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}](https://tex.z-dn.net/?f=r%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3D18%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B18%7D)
<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>
Then, we calculate the arguments of the trigonometric functions:

We can now calculate for each value of k:
![\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D0%5Ccolon%20%5C%5C%20z_0%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B0%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B0%7D%7B2%7D%29%29%29%20%5C%5C%20z_0%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B%5Cpi%7D%7B8%7D%29%20%5C%5C%20z_0%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%20e%5E%7Bi%5Cfrac%7B%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D1%5Ccolon%20%5C%5C%20z_1%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B1%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B1%7D%7B2%7D%29%29%29%20%5C%5C%20z_1%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B5%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B5%5Cpi%7D%7B8%7D%29%29%20%5C%5C%20z_1%3D%5Csqrt%5B4%5D%7B18%7De%5E%7Bi%5Cfrac%7B5%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D2%5Ccolon%20%5C%5C%20z_2%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B2%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B2%7D%7B2%7D%29%29%29%20%5C%5C%20z_2%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B9%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B9%5Cpi%7D%7B8%7D%29%29%20%5C%5C%20z_2%3D%5Csqrt%5B4%5D%7B18%7De%5E%7Bi%5Cfrac%7B9%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D3%5Ccolon%20%5C%5C%20z_3%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B3%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B3%7D%7B2%7D%29%29%29%20%5C%5C%20z_3%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B13%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B13%5Cpi%7D%7B8%7D%29%29%20%5C%5C%20z_3%3D%5Csqrt%5B4%5D%7B18%7De%5E%7Bi%5Cfrac%7B13%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
Answer:
The four roots in exponential form are
z0 = 18^(1/4)*e^(i*π/8)
z1 = 18^(1/4)*e^(i*5π/8)
z2 = 18^(1/4)*e^(i*9π/8)
z3 = 18^(1/4)*e^(i*13π/8)
Answer:
D) 0.42 + 0.17 x = 0.93
Step-by-step explanation:
0.93 - 0.42 = 0.51
0.51 / 0.17 = 3
3 ounces and the answer is D