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3 years ago

Solve simultaneous equation 2x + 4y = 1 3x-5y=7

1 answer:

5 0

$\left \{ {{2x+4y=1\ \ | *(-3)} \atop {3x-5y=7\ \ | *2}} \right. \\\\
 \left \{ {{-6x-12y=-3} \atop {6x-10y=14}} \right.\\+----\\add\ both\ equations\\\\
-22y=11\ \ \ | divide\ by\ -22\\\\
y=-\frac{1}{2}\\\\
2x=1-4y\ \ | divide\ by\ 2\\\\
x=\frac{1-4y}{2}=\frac{1-4*(-\frac{1}{2})}{2} =\frac{3}{2}$

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WILL GOVE BRAINIEST Kim and Tegan train for the iron man on the weekend. They bike at a constant rate of miles per hour. The tab

Rus_ich [418]

Answer:

C 210

Step-by-step explanation:

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(edit: to solve for Kim's missing values, you do the same with 2 and 32)

6 0

3 years ago

Assume z = x + iy, then find a complex number z satisfying the given equation. d. 2z8 – 2z4 + 1 = 0

kodGreya [7K]

Answer: complex equations has n number of solutions, been n the equation degree. In this case:

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}$

Step-by-step explanation:

I start with a variable substitution:

$Z^{4} = X$

Then:

$2X^{2}-2X+1=0$

Solving the quadratic equation:

$X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}$

$X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.$

Replacing for the original variable:

$Z=\sqrt[4]{0,5+0,5i}$

or $Z=\sqrt[4]{0,5-0,5i}$

Remembering that complex numbers can be written as:

$Z=a+ib=|Z|e^{ic}$

Using this:

$Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.$

Solving for the modulus and the angle:

$Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.$