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Flashcards by LIDA KATERINE RAMIREZ RAMIREZ , updated more than 1 year ago
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| Question | Answer |
| Quadratic Formula - solve: \(a\)\(x^2\)+\(b\)\(x\)+\(c\)=\(0\) where \(a\) \(\neq\) \(0\) | \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}\\ \end{array} |
| Circumference of a Circle: | \(2\)\(\pi\)\(r\) or \(\pi\)\(d\) where \(r\)=radius, \(d\)=diameter |
| Area of a Circle: | \(\pi\)\(r\)\(^2\) |
| Pythagoras theorem In any right-angled triangle where \(a\), \(b\) and \(c\) are the length of the sides and \(c\) is the hypotenuse: | \(a^2\)+\(b^2\)=\(c^2\) |
| Trig: In any right-angled triangle \(ABC\) where \(a\), \(b\) and \(c\) are the length of the sides and \(c\) is the hypotenuse: \(sinA\)= | \(sinA\)=\(\frac{a}{c}\) |
| Trig: In any right-angled triangle \(ABC\) where \(a\), \(b\) and \(c\) are the length of the sides and \(c\) is the hypotenuse: \(cosA\)= | \(cosA\)=\(\frac{b}{c}\) |
| Trig: In any right-angled triangle \(ABC\) where \(a\), \(b\) and \(c\) are the length of the sides and \(c\) is the hypotenuse: \(tanA\)= | \(tanA\)=\(\frac{a}{b}\) |
| Sine Rule: | \(\frac{a}{sinA}\)=\(\frac{b}{sinB}\)=\(\frac{c}{sinC}\) |
| Cosine Rule: | \(a^2\)= \(b^2\)+\(c^2\)-\(2\)\(b\)\(c\) \(cosA\) |
| Trigonometry: Area of a Triangle | \(\frac{1}{2}\)\(a\)\(b\)\(SinC\) |
| Area of a Trapezium= (Where \(a\) and \(b\) are the lengths of the parallel sides and \(h\) is their perpendicular separation) | \(\frac{1}{2}\) (\(a\) + \(b\))\(h\) |
| Volume of a Prism: | area of cross section × length |
| Compound interest: Where \(P\) is the principal amount, \(r\) is the interest rate over a given period and \(n\) is number of times that the interest is compounded, Total accrued= | Total accrued= \begin{array}\(P\left(1+ \frac{r}{100}\right)^n\end{array} |
| Where P(A) is the probability of outcome A and P(B) is the probability of outcome B: P(A or B) = | P(A or B) = P(A) +P(B) - P(A and B) |
| Where P(A) is the probability of outcome A and P(B) is the probability of outcome B: P(A and B) | P(A and B) = P(A given B) P(B) |
| Curved surface area of a cone: | \(\pi\)\(r\)\(l\) |
| Surface area of a Sphere: | \(4\)\(\pi\)\(r\)\(^2\) |
| Volume of a Sphere: | \(\frac{4}{3}\)\(\pi\)\(r\)\(^3\) |
| Volume of a Cone: | \(\frac{1}{3}\)\(\pi\)\(r\)\(^2\)\(h\) |
| Final Velocity \(v\): | \(v\)=\(u\)+\(at\) (\(u\)=initial velocity, \(a\)=constant acceleration, \(t\)=time taken) |
| Displacement \(s\): | \(s\)=\(ut\) +\(\frac{1}{2}\)\(a\)\(t\)\(^2\) (\(u\)=initial velocity, \(a\)=constant acceleration, \(t\)=time taken) |
| Velocity \(v\)\(^2\): | \(v\)\(^2\)=\(u\)\(^2\)+ \(2\)\(as\) (\(u\)=initial velocity, \(a\)=constant acceleration, \(s\)=displacement) |
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