Answer:
When A and B are not both zero, the graph of Ax + By = C is always a line. The graph of Ax + By = C, where A and B are not both zero, is a line. Divide both sides by B. Because the form Ax + By = C can describe any line, it is called the standard form of an equation for a line.
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Answer:
Step-by-step explanation:
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
so Sin(x) = d/f
Cos(x) = e/f
Tan(x) = d/e
got it? :)
let's firstly convert the mixed fractions to improper fractions, and then add.
![\bf \stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}~\hfill \stackrel{mixed}{2\frac{5}{6}}\implies \cfrac{2\cdot 6+5}{6}\implies \stackrel{improper}{\cfrac{17}{6}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{13}{4}+\cfrac{17}{6}\implies \stackrel{\textit{we'll use the LCD of 12}}{\cfrac{(3)13~~+~~(2)17}{12}}\implies \cfrac{39+34}{12}\implies \cfrac{73}{12}\implies 6\frac{1}{12}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%204%2B1%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B13%7D%7B4%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B5%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%206%2B5%7D%7B6%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B17%7D%7B6%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ccfrac%7B13%7D%7B4%7D%2B%5Ccfrac%7B17%7D%7B6%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20the%20LCD%20of%2012%7D%7D%7B%5Ccfrac%7B%283%2913~~%2B~~%282%2917%7D%7B12%7D%7D%5Cimplies%20%5Ccfrac%7B39%2B34%7D%7B12%7D%5Cimplies%20%5Ccfrac%7B73%7D%7B12%7D%5Cimplies%206%5Cfrac%7B1%7D%7B12%7D)
Answer:
Step-by-step explanation:
A = LW + ½πr²
A = 15(12) + ½(3.14)(12²/4)
A = 180 + 56.52
A = 236.52 m²
Answer:

Step-by-step explanation:
<u>Biquadratic Equations</u>
Solve:

The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.
Solving such equations requires to express the equation as a second-degree equation with
as the variable.
Rewriting the equation:

The quadratic equation can be factored as:

It leads to two equations:


The first equation has imaginary roots. Solving for x:




Where

The second equation has two real roots:





The roots are:
