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Trigonometry help please?
If cos B is 5/13, find sin B, tan B, sec B, and cot B?sin= y/r= 12/13cos= x/r= 5/13tan= y/x= 12/5csc= r/y= 13/12sec= r/x= 13/5cot= x/y= 5/12x2+y2=r2
Simplify this expression 13 -12--5 a -6 b -30 c 6 d 30?
30.. B
How can you show that the modular equation 22x41(mod13) has a unique solution and find that solution?
Let 22*41 = 13*a + b where a and b are integers and 0
If 27 equals 13 divided by 37 times 27 plus b what does b equal?
27 = ((13/37) * 27) + b b = 17,51or27 = (13/(37*27)) + bb = 26,987or27 = (13/((37*27)+b))b = -998,52
2a-b equals 17 3a plus 4b equals -13?
2a-b equals 17 3a plus 4b equals -13? Elimination Method If you can multiply one of the equations by a number and eliminate one of the variables, that is an easy to find the value of the other variable. I see -b in Eq. #1 and +4b in Eq. #2. I can multiply by Eq. #1 by 4 and add to Eq. #2 eliminate b. Eq. #1 = 2a - b = 17 Eq. #2 = 3a + 4b = -13 4* (2a - b = 17) = 8a - 4b = 68 Now add 4* Eq. #1 to Eq. #2 ..8a - 4b = 68 +.3a + 4b = -13 .11a + 0b = 55 11a = 55 divide both sides by 11 a = 5 Substituting a = 5 into Eq. #1 (2 * 5) - b = 17 10 - b = 17 subtract 10 from both sides - b = 17 - 10 -b = 7 Divide by -1 b = -7 Check in Eq. #2, by substituting a = 5 and b = -7 3a + 4b = -13 (3 * 5) + (4 * -7) = -13 15 + -28 = -13 -13 = -13 The other way of solving simultaneous equations is Substitution. Solve for one variable in terms of the other variable. Eq. #1 = 2a - b = 17 Eq. #2 = 3a + 4b = -13 Solve Eq.#1 for a in terms of b. 2a - b = 17 add +b to both sides 2a = b + 17 divide (b + 17) by 2 a = (b + 17) ÷ 2 Substitute a = (b + 17) ÷ 2 into Eq.#2, and solve for b Eq. #2 = 3a + 4b = -13 3 * [(b + 17) ÷ 2] + 4b = -13 Multiply 3 * (b + 17) (3b + 51)÷2 + 4b = -13 Multiply both sides by 2 to eliminate the (÷2) (3b + 51) + 8b = -26 Add (3b + 8b) 11b + 51 = -26 Add -51 to both sides 11b = -77 Divide both sides by 11 b = -7 Substitute b = -7 into Eq.#1 Eq. #1 = 2a - b = 17 2a + 7 = 17 Subtract 7 from both sides 2a = 10 Divide both sides by 2 a = 5 That Substitution sure was complicated compared to the Elimination method. I wonder if it would have been easier to solve Eq.#1 for a in terms of b. That [(b + 17) ÷ 2] was the trouble maker. Eq. #1 = 2a - b = 17 Add b to both sides 2a = 17 + b subtract 17 from both sides 2a -17 = b reverse sides b = 2a - 17 Substitute (b = 2a - 17) into Eq.#2, and solve for a Eq. #2 = 3a + 4b = -13 3a + 4 * (2a - 17) = -13 Multiply 4 * (2a - 17) 3a + (8a - 68) = -13 Add 3a + 8a 11a - 68 = -13 Add +68 to both sides 11a = 55 Divide by 11 a = 5 Eq. #1 = 2a - b = 17 Substitute a = 5 into Eq.#1 2 * 5 - b = 17 Multiply 2 * 5 = 10 10 - b = 17 Add -10 to both sides b = -7 We got the same answers as we did by using the elimination method, but the substitution was more work. Sometimes using the elimination method, you have to multiply both equations by 2 different numbers to eliminate a variable, but I still think it is easier.
