I need to solve for the three unknowns a, b, and c in the following equations. 3a - 5c = -9, 2a + 7b = 11, a + b + c = 6. The answer is (2, 1, 3), but how do I get there?
First of all, check the equations. It has 3 variables (A, B, C) and 3 independent equations. So it's solvable. (If it's not solvable, then we cannot have a exact answer.)Before doing any calculate, exam each equation. In the first one, it tells the relationship between A and C. The second one, A and B. The third one, A , B, C. It give us a hint. We can use A to express B or C. B=(11-2A)/7C=(3A+9)/5Then put them into the third equation.A+ (11-2A)/7 +(3A+9)/5 =6Only ONE variable, A, in this equation. We can get A form this one.Multiply 35 to both side of the equation. (35 is the least common multiple for 7 and 5)We got, 35A + 55 - 10A +11A +63 = 210Then we get A=2.And, B=(11-2A)/7 = 1C=(3A+9)/5 = 3
Let ln2=a, ln3=b, and ln5=c. Express the following logarithms in terms of abc? ln2000= ln14.4= ln.075= log(?
1.) ln(2000) = ln(2^4 * 5^3) = ln(2^4) + ln(5^3) = 4*ln(2) + 3*ln(5) = 4a + 3c. 2.) ln(14.4) = ln(72/5) = ln(2^3 * 3^2 / 5) = ln(2^3) + ln(3^2) - ln(5) = 3*ln(2) + 2*ln(3) - ln(5) = 3a + 2b - c. 3.) ln(0.075) = ln(75/1000) = ln(3/40) = ln(3/(2^3 * 5)) = ln(3) - ln(2^3) - ln(5) = ln(3) - 3*ln(2) - ln(5) = b - 3a - c. 4.) log(base 2)(600) = log(base e)(600) / log(base e)(2) by the change of base rule, so log(base 2)(600) = ln(600) / ln(2) = ln(2^3 * 3 * 5^2) / ln(2) = [ln(2^3) + ln(3) + ln(5^2)] / ln(2) = [3*ln(2) + ln(3) + 2*ln(5)] / ln(2) = [3a + b + 2c]/a. Note that part of the 'trick' here is to know how to manipulate powers of 2, 3 and 5 to get 2000, 14.4, 0.075 and 600. With 2000 and 600 you just need to find the prime factorization. For 14.4 we note first that 5*14.4 will be a whole number, namely 72. Then since 72 = 8*9 = 2^3 * 3^2 we can proceed with the natural logarithm calculation. With 0.075, we first note that 1000*0.075 = 75, so that 0.075 = 75 / 1000 = 3*25 / (40*25) = 3/40 = 3/(5* 2^3), from which we can proceed with the natural log calculation.
Solve An Easy Factoring Question! FAST 10 POINTS!?
If A + B = 3, then A = 3 - B and (3 - B)^2 + B^2 = 7 9 - 6B + B^2 + B^2 = 7 2B^2 - 6B + 9 = 7 2B^2 - 6B - 2 = 0 (divide both sides by 2) B^2 - 3B - 1 = 0 B= 3.302775637731995, -.302775637731995 (These numbers are A & B) 3.302775637731995^4 + (-.302775637731995)^4 = 119
If a3 + b3 + c3 = 3abc and a,b,c are positive numbers, then prove that a = b = c.?
a3 + b3 + c3 = 3abc a3 + b3 + c3 - 3abc = 0 (a + b + c )(a2 + b2 + c2 -ab - ac -bc) = 0 here (a + b + c) cannot be equal to zero since a,b,c are positive. Therefore (a2 + b2 + c2 -ab - ac -bc) = 0 2a² + 2b² + 2c² -2ab - 2ac - 2bc = 0 a² + b² -2ab + a² +b² +2c² - 2ac -2bc = 0 (a-b)² + a² + c² -2ac + b² + c² -2bc = 0 (a-b)² + (a-c)² + (b-c)² = 0 (a-b)² , (a-c)² ,(b-c)² >=0 since postive numbers cannot be added to make zero, each bracket should be zero a-b = 0 a-c = 0 b-c =0 a = b a= c b = c a = b = c
If log[a+b/3]=1/2[loga +logb] prove that a/b+b/a=7?
log[a+b/3]=1/2[loga+logb] 2log[a+b/3]=log[ab] log[a+b/3]^2=log[ab] [a+b]^2/9=ab a^2+b^2+2ab=9ab a^2+b^2=7ab dividing by ab a/b+b/a=7
If (2a+b)/(a+4b)=3, then what will be the value of (a+b)/(a+2b)?
(2a+b)/(a+4b) = 3Multiply both sides by the LCD, (a+4b), to eliminate the denominator:2a+b = 3(a+4b)distribute the 3:2a+b = 3a + 12bSubtract 3a from both sides:-a +b = 12bSubtract b from both sides, and then divide by -1:-a = 11ba = -11bSince we’ve solved for a we can now substitute for a and solve in terms of b:(2a+b)/(a+4b) = 3a = -11b(2*-11b+b)/(-11b+4b) = 3(-22b+b)/ (-7b) = 3(-21b)/(-7b) = 3Divide; the gcf is -7b:3 = 3So b = 3. This means a = -11 * 3 = -33Substitute for both a and b into the original equation to verify both solutions are true:(2a+b)/(a+4b) = 3a = -33b = 3(2*-33+3)/(-33+4*3) = 3(-66+3)/(-33+12) = 3(-63)/(-21) = 3Reduce the fraction; the GCF is -21:3 = 3so a = -11 and b = 3.
If a+b=6 and a-b=4 then what is the value of a²+b²?
One quick way to solve this will bea+ b =6(a+b)^2 = 36. ….(1)a- b = 4(a-b)^2 = 16…….(2)Adding (1) & (2) we get,(a+b)^2 + (a-b)^2=a^2+ b^2 +2ab +a^2+b^2–2ab =522(a^2+b^2) = 52a^2+b^2 = 26Alternate traditional waya+ b =6 …….(1)a- b = 4 …….(2)Subtracting (1) & (2) we get,2b = 6–4b = 1Substituting in (1) we get,1+ a = 6a = 5Therefore a^2+ b^2 =1*1 + 5*5 = 1+ 25 = 26
What is the value of A+B if tan A=1/2, cot B=3?
tanA=1/2cotB=3according to trigonometry ruletanB= 1/cotBthantan B= 1/3apply the trigonometry formula of tan(A+B)tan(A+B)= tanA+tanB/1- tanA*tanBtan(A+B)=(1/2+1/3)/(1–1/3*1/2)- eq1after solve the eq1 then gettan(A+B)=(5/6)/(5/6)both are cancel out then givetan(A+B)=1 -eq2after solution i move the tan in right hand side then tan is change to arctan(tan inverse)A+B= arctan 1arctan 1= Pi/4 or 45 degreevalue of A+B= pi/4 or 45 degree