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MathHow do you solve an equation for a given variable? For example, how would you do: 5(2m+3) (12m)...5(2m+3)  (12m) = 2[3(3+2m)  (3m)] First we will expand brackets starting from the inside out: 5*2m + 5*3 1 +2m = 2[3*3+3*2m 3 +m] 10m + 15 1 +2m = 2[9 +6m 3 +m] Group similar terms: 12m +...

Math5n + 1 = 5n 1\ First you need to group similar terms together. In other words, group n terms together on one side, and the constant terms on the other sides: 5n  5n = 11 0 = 2 Then the...

Mathf(x) = e^x / (x+2) Let f(x) = u/v such that: u= e^x ==> u'= e^x v= (x+2) ==> v' = 1 Noe by definition: f'(x) = (u'vuv')/v^2 =[ (e^x)*(x+2)  (e^x)*1]/(x+2)^2 = (e^x)(x+2...

Math3, a, b, 192 are parts of a geometric progression: a1= 3 a2= 3*r = a a3= 3*r^2 = b a4= 3*r^3 = 192 ==> 3r^3 = 192 Divide by 3: ==> r^3 = 192 /3 = 64 ==> r^3 = 64 Now take the cubic root...

Mathf(x) = x^2 + mx 15 Since x=3 is one of the roots for f(x) , then f(3) =0 Let us substitute: f(3) = 3^2 + m*3  15 = 0 ==> 9 +3m 15 = 0 ==> 3m 6 = 0 ==> 3m = 6 ==> m = 6/3 = 2 ==>...

Mathln e^(x+2) = 5 + ln e^2 We know that: ln e^a = aln e = a*1 = a ==> ln e^(x+2) = 5 + ln e^2 ==> (x+2) = 5 + 2 ==> x+2 = 7 ==> x= 72 ==> x= 5

Mathlog 2 (2x) + log 4 (4x^2) = 3 First let us rewrite : We know that: log a b = log c b/log c a ==> log 4 (4x^2) = log 2 (4x^2) / log 2 4 But log 2 4 = 2 ==> log 4 (4x^2) =(1/2) log 2 (4x^2) =...

Math7 + 5(x2) = 3(5x+1) 12 First let us open brackets: ==> 7 + 5x  10 = 15x +3 12 Simplify: ==> 5x 3 = 15x  9 Now group similar terms: ==> 10x = 6 Now divide by 10: ==> x = 10/6= 5/3...

Mathcosx =23x^2+0.5 To solve for x. Solution: Let x be in radians and f(x) = cosx +0.23x^20.5 . Now f'(x) = sinx +2(0.23)x = sinx+0.46x. Setting f'(x) = 0 we get x= 0 f"(x) = (sinx+0.46)' = ...

Math(3/7) t + 4 = (1/7) t  6 First let us multiply by 7: 7*(3/7)t + 4*7 = 7*(1/7)t  6*7 ==> 3 t + 28 = t  42 Now group similar: ==> 3t  t = 42  28 ==> 2t = 70 ==> t = 70/2 = 35...

Math8x  5(2+x) = 2(x+1) First you need to open brackets: 8x  5*2 5*x = 2*x + 2*1 8x  10  5x = 2x + 2 Now group similar: 8x  5x  2x 10  2=0 x  12 = 0 Now add 12 to both sides: ==> x = 12 To...

Mathlog 2 (3+log 3x) = 2 ==> 3+ log 3x = 2^2 ==> 3+ log 3 x = 4 ==> log 3 x = 1 ==> x = 3^1 = 3 ==> x= 3 Now let us check: log 2 (3+ log 3 (x) = 2 loh 2 (3 + log 3 (3) = 2 We know that...

MathLet the first root be z such that: z= (1i)(2+i) = 2 +i 2i i^2 = 3 i The the first root is: z= 3i The second root would be z' such that: z'= 3+i Now to find the equation: z+z' = 3i + 3 +i...

Mathz+i = 2(1z') Let z = a+bi ==> z'= abi Let us substitute: ==> (a+bi) +i= 2(1 (abi) ==> a+bi +i= 2(1a + bi) ==> a+(b+1)i = 22a + 2bi ==> a + (b+1)i  2 + 2a  2bi= 0 ==> 3a +...

Math(1+i)(13i) Let us expand brackets: (1+i)(13i) = 1*1 + 1*3i + i*1 + i*3i = 1 3i + i 3i^2 But we know that i^2 = 1 ==> (1+i)(13i) = 1  3i +i 3*1...

Mathx+my 5 = 0 (1,5) is on the line. Then the point (1,5) should verify the equation. Let us aubstitute: x+ my  5 = 0 1+ m(5) 5 = 0 1+ 5m  5 = 0 5m  4 = 0 5m = 4 ==> m = 4/5 Then the equation...

MathTo determine the coordinates of the intercepting point of the graphs of f and g, we'll have to solve the system: y = x2 (1) y = 3x+4 (2) x2 = 3x+4 We'll isolate x to the left side: x3x = 4+2 2x...

MathIf we'll substitute infinite in the expression of the limit, we'll have the indetermination, type 0*infinite. We'll rewrite the expression of the limit: lim (1/n)*ln(1+e^n) = lim ln (1+e^n)^(1/n)...

Math7x + 2y = 12 ....(1) x3y = 5 ............(2) Using the substotitution methodrewrite (2): ==> x= 3y + 5 Now substotitute in (1): 7x + 2y = 12 7(3y+5) + 2y = 12 21y + 35 = 2y = 12 23y = 23...

MathLet S = 1/1*2 + 1/2*3 + ...+ 1/2010*2011 Now let us rewire: We know that 1/n(n+1) = 1/n 1/(n+1) Then: 1/1*2 = 1  1/2 1/2*3 = 1/2  1/3 ...... 1/2010*2011 = 1/2010  1/2011 S = 1/1  1/2 + 1/2 ...

Mathf = sum k*(k+1)/3x*(x+1)(x+2) S = k(k+1= k^2 + k= Sk^2 + sK Sk^2 = 1 + 4 + 9 + 16 + ...+n = n(n+1)(2n+1)/6 Sk = 1+ 2 + 3+ ...+ n = n(n+1)/2 Then: S = n(n+1)(2n+1)/6 + n(n+1)/2 = [n(n+1)(2n+1) +...

Mathlim (1+2+...+n)/(n^2 + 3n + 1) n > inf By substituttion: lim (1+2+ ..+n)/n^2 + 3n + 1) = inf/inf But we know that: 1+ 2+ ..._+ n = n(n+1)/2 ==> lim n(n+1)/2(n^2 + 3n + 1) ==> lim (n^2 +...

MathLet the numbers be: a, b, c Then: (a+b+c)/3 = 23 ==> a+ b + c = 69 .......(1) But we know that: b = a+ 6 ==> a = b6 .........(2) also: b+c = 54.........(3) Now let us substitute (2) and (3)...

Matha= (111/2)(112/2)(113/2)...(1126/2) l m3l = a Let us write some more terms: a= (111/2)(112/2) ....(1122/2).....(1126/2) = (111/2)(112/2).....(0)........(1126/2) ==> a = 0 Now to...

MathWe'll substitute infinite into the given expression: lim n^2*sin(1/n)/(n+1) = inf. * sin (1/inf.) / (inf. + 1) = inf.*sin0/inf. lim n^2*sin(1/n)/(n+1) = 0*inf. As we can see, this is a case of...

MathTo solve the equation, we'll consider 4 cases of study. The expressions 3x4 and 34x can be either positive or negative. Case 1: 3x4 = 34x Case 2: (3x4) = 34x Case 3: 3x4 = (34x) Case 4:...

Mathcosx/ (1sinx) = 1+sinx Cross multiply: ==> cosx = (1+sinx)*(1sinx) ==> cosx = 1  sin^2 x But we know that: cos^2 x = 1sin^2 x ==> cosx = cos^2 x ==> cos^2 x  cosx = 0 ==>...

MathWe'll differentiate the composed function f(x) noticing that it contains a product and the logarithm of a quotient: (u*v)' = u'*v + u*v' f' = {sin (x^2 + x + 5)*ln[(2x+1)/(2x1)]}' f' = [sin (x^2...

Math(3x+2)^1/3  (x1)^1/3 = 1 Let us cube both sides: [(3x+2)^1/3)  (x1)^1/3]^3 = 1 But we know that: (ab)^3 = a^3  b^3  3ab(ab) ==> [(3x+2)^1/3  (x1)^1/3]^3 = (3x+2)  (x1) ...

MathFor the beginning, let's recall that the terms arctan (1/3) and arctan [(5sqrt36)/3] are angles. Let's note arctan (1/3) = a and arctan [(5sqrt36)/3] = b We'll apply the tangent function to the...

MathWe'll complete the square in the given expression: (sin x)^4+(cos x)^4 + 2(sinx*cosx)^2  2(sinx*cosx)^2 =1/2 [(sin x)^2+(cos x)^2]^2  2(sinx*cosx)^2 =1/2 But, from the fundamental fromula of...

MathWe'll substitute arctan x = t We'll rewrite the equation in t: 2/t  t = 1 We'll multiply by t the terms without denominators: 2  t^2 = t We'll move all terms to the right side: t^2 + t  2 = 0...

MathTo determine x, we'll rewrite the equation so that to caontain only the function cosine. We'll apply the fundamental formula of trigonometry: (sinx)^2 + (cos x)^2 = 1 (sinx)^2 = 1(cos x)^2 We'll...

MathWe could also write: sin x + sin 2x = sin 3x We'll substitute sin 2x and sin 3x by the followings: sin 2x = sin (x+x) = sin x*cos x+cos x*sin x sin 2x = 2sinx*cosx sin 3x = sin (2x+x) = sin2x*cosx...

Mathln (2^x1), ln (2^x + 3) , ln (2^x + 5) ==> ln (2^x +3)  ln (2^x 1) = ln (2^x + 5)  ln (2^x +3) But we know that: ln a  ln b = ln a/b ==> ln (2^x +3)/(2^x 1) = ln (2^x +5)/(2^x +3)...

Mathan = n^2 + n + 1 Let us calculate some terms in the progression: a1= 1^2 + 1 + 1 = 3 a2 = 2^2 + 2 +1 = 7 a3= 3^2 + 3 + 1 = 13 a4= 4^2 + 4 + 1 = 21 We notice that : a2a1 , a3a2, and a4a3 are not...

Math3, 9, 27, ..... We notice that what we have is a geometric progression where r= 3. a1= 3 a2= 3*3 = 9 a3= 3*3^2 = 27 Then: an = a1*3^(n1) ...... a11= 3*3^10 = 3^11 = 177147. an = 3*3^(n1) = 3^n

Mathsina + cosa = 1 ==> sina = 1cosa We know that: tan2a = 2tana / [1(tana)^2] But tana = sina/cosa ==> tan2a = 2(sina/cosa) / [1 (sina/cosa)^2] = 2(1cosa)/cosa /...

Math1/(1i) + 1/(1+i) First let us rewrite using the common denominator: [(1+i) + (1i)]/(1i)(1+i) Now open brackets: ==? (1+i+1i)/(1^2  i^2) ==> 2/1+1 = 2/2 = 1 ==> z = 1 + 0i Then the real...

Mathsqrt(x^2 + 36x + 36)  6*sqrt(x1) = 0 Let us move 6sqrt(x1) to the right sides: ==> sqrt(x^2 + 36x + 36) = 6*sqrt(x1) =Now ley us square both sides to get rid of the sqrt: ==> x^2 + 36x +...

Math5^x^2  5^(5x+6) = 0 ==> 5^x^2 = 5^(5x + 6) ==> x^2 = 5x + 6 ==> x^2 +5x 6 = 0 ==> (x+6)(x1) = 0 ==> x1= 6 ==> x2= 1

Mathy = e^(2x) . sin 5x Let y = u*v such that: u = e^2x ==> u' = 2e^2x v = sin5x ==> v' = 5cos5x ==> y'= u'v + uv' = (2e^2x)*sin5x + (e^2x)*5cos5x = 2e^2x...

Math2, a, b , 17 are terms is an arithmetical progression. The: a1= 2 a2= a1+ r = 2 + r = a a3= a1+ 2r = 2 + 2r = b a4= a1+ 3r = 2+ 3r = 17 ==> 3r = 15 ==> r = 5 ==> a= 2+r = 2+5 = 7 ==> b...

Mathlog 2x  log (x5) = log 5 We know that: log a  log b = log a/b ==> log (2x/(x5)= log 5 ==> 2x/(x5) = 5 Now cross multiply: ==> 2x = 5(x5) ==> 2x = 5x  25 ==> 3x = 25 ==> x =...

MathArea of the rectangle is: a = length * width = L*w But we know that: L = 2+ 2w Also: 2L + 2W = 22 Now substitute with L value: ==> 2L + 2W = 22 ==> 2(2+2w) + 2w = 22 ==> 4 + 4w + 2w = 22...

Mathx+2y =3 ..........(1) 4x + 5y = 6........(2) Let us use the substitution method: from (1): x = 3 2y Now substitute in (2): 4x + 5y = 6 4(32y) + 5y = 6 12  8y + 5y = 6 3y = 6 y= 2 x = 32y = 3...

Math5/(x1) + 2x/(x^21) Let us simplify the denominator: We know that: x^2 1 = (x1)(x+1) ==> 5/(x1) + 2x/(x1)(x+1) ==> 5(x+1)/(x1)(x+1) + 2x/(x1)(x+1) ==> (5x + 5 + 2x)/(x1)(x+1)...

MathTo prove that a function is invertible, we have to demonstrate that the function is a bijection. If a function is both injection and surjection, then the function is a bijection. In order to verify...

MathThe equation for the ine is: y y1 = m(xx1) m= (y2y1)/(x2x1) = (42)/(31) = 1 ==> y2 = 1(x1) ==> y= x  1 + 2 ==> y= x+1 ==> x + y 1 = 0 The distance between the line and the...

Mathf= 2x/(x+1)(x^2 +1) F = aln (1+x) + b ln(1+x^2) +c*arctanx F' = a/(1+x) +2bx/(1+x^2) + c/(1+x^2) = [a(1+x^2) + 2bx(1+x) + c(1+x)]/(1+x^2)(1+x) = ( a + ax^2 + 2bx + 2bx^2 + c + x...