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ACT Practice Trigonometry Area of an oblique triangle {Trigonometry} {given two sides and an included angle} Area = π ππ π¬π’π§ π π Area of βΏπππ 1 30 = sin 70° xy 2 30 = 1 .93969 xy 2 60 = .93969xy xy = 63.85 = π ππ π¬π’π§ π π = π ππ π¬π’π§ π π Area of βΏπππ 1 A = sin 110° xy 2 A= 1 .93969 xy 2 Areas are proven to be equal using the property stated above!!! © Mr. Sims ACT Practice Trigonometry c a b c² = a² + b² β 2ab cos β C {given law of cosines} a² = b² + c² β 2bc cos β A {put in terms of side a} a² = 12² + 18² β 2 12 18 cos 40° {substituted values in} a = 12² + 18² β 2 12 18 cos 40° {square root of each side} © Mr. Sims ACT Practice Intermediate Algebra The nth term of an arithmetic sequence can be represented by: an = a + (n β 1)βd an is the nth term a is the first term of the sequence n is the number of term d is the common difference 8, ___ , ___ , ___ , 13 10th term 6th term the 10th term exceeds the 6th term by 5 10th term equals 6th term plus 5 a + (10 β 1)βd = a + (6 β 1)βd + 5 {the 10th term is 5 more than the 6th term} 9d = 5d + 5 {subtracted a from each side and worked inside parentheses} 4d = 5 {subtracted 5d from each side} d = 1.25 {divided each side by 4} the common difference is 1.25 Work backwards to find the first 4 terms: 8 β 1.25 = 6.75 5th term 6.75 β 1.25 = 5.5 {4th term} st four terms together Add the 1 rd 5.5 β 1.25 = 4.25 {3 term} = 5.5 + 4.25 + 3 + 1.75 4.25 β 1.25 = 3 {2nd term} 3 β 1.25 = 1.75 {1st term} = 14.5 © Mr. Sims ACT Practice Elementary Algebra Work backwards from the solution to create a quadratic equation: x = -3 or x = -3 {the graph crosses the x-axis, twice, at -3} x + 3 = 0 or x + 3 = 0 {added 3 to both sides of each equation} (x + 3)(x + 3) = 0 {both can be multiplied to equal 0} x2 + 6x + 9 = 0 {squared the binomial} x2 + mx + n = 0 © Mr. Sims ACT Practice Intermediate Algebra 5 units from -3 would be -8 and also 2 Absolute value is the distance away from zero. |x + 3| = 5 {means (x + 3) is 5 units from zero} x + 3 = 5 or x + 3 = -5 {(x + 3) could be 5 or -5} x=2 or x = -8 {subtracted 3 from both sides of each equation} The solution set to |x + 3| = 5 is the set of real numbers that are 5 units from -3 © Mr. Sims Any rebroadcast, reproduction, modification or other use of the work, presentations, and materials from this site without the express written consent of Mr. Sims, is prohibited. © Mr. Sims. All rights reserved. © Mr. Sims