Can someone help me solve c^2 + 12c + 32 = 0?
c^2 + 12c + 32 = 0 Or (c + 8)(c + 4) = 0 Or c = --8 or c = --4.
Could you help me solve my problem?
Siddarth nailed it!No, nobody can do that but yourself! Only you know what the problem really is, only you can look into yourself and find the answer, only you can decide if you need a specific helping hand from somebody else (but you know exactly what that is, and it’s not solving “all” your problem!), only you can take the decision to do so, for if you want to solve the problem but don’t take the decison to do so, nobody in the world can do it for you!So, look at the problem, find the solution and get going. If it doesn’t work look at the problem again and start all over, etc., etc., until your problem is solved.Remember that all thousand mile trips start with but one step…so get going and take that step!
Can you help me to find the solve?
We hope that you learn methods for finding solutions; do not just copy. e) The graph of this function looks like this https://www.wolframalpha.com/input/?i=y+... a) y = x^3 + 3x^2 - 9x - 2 The y intercept is where x = 0 so it is y = -2 The graph suggests that x = 2 is one of the x intercepts, and factoring, x^3 + 3x^2 - 9x – 2 = (x – 2)(x^2 + 5x + 1) The other two intecepts are [-5 ±√(21)]/2, or roughly x ~ -0.208712 and x ~ -4.79129 b) dy/dx = 3x^2 + 6x – 9 = 3(x – 1)(x + 3) = 0 at turning points Graph indicates max at x = -3, min at x = +1 c) Between x = -3 and x = +1, dy/dx is negative When x < -3 and where x > 1, is positive d) The 2nd derivative measures concavity, up or down, so, an inflection point is where f’’ = 6x + 6 = 0, that is where x = -1.
Can anyone help me solve ln2x+ln(x-5)=ln8?
ln(2x)+ln(x-5)=ln(8) ln(2x(x-5))=ln(8) 2x(x-5)=8 2x^2 - 10x - 8 = 0 x^2 - 5x - 4 = 0 x = [5 ± sqrt(25 - 4(-4))]/2 x = 5/2 ± sqrt(41)/2 x ≈ {-0.702, 5.702} For ln(2x) to be real number 2x>0, x>0 x = 2.5 + 0.5sqrt(41) ≈ 5.702
HELP ME!!!!how do i solve this math problem?
The total annual interest from two bank accounts is $481. One account earns 5.9% annual interest and the other earns 6.75%. If the two accounts contain a total of $7,600, how much is the account with the lower interest?!! please solve/answer the problem...
Help me solve this equation.?
the perimeter of a rectangular corn field is 628 m. THe length of the field is 6 m greater than the width. find the area of the field. A=lw and P= 2l+2w I know L= X+6 and W= X but how do I solve this.
Can someone help me solve this integral?
Note that [math]\int dx/(ax+b) = \log(ax+b)/a[/math] and is not rational, whereas for [math] n \geq 2, n\in \mathbb{N} [/math], [math] \int dx/(ax+b)^n [/math] is rational.When you try to split the integrand in your problem into partial fractions, you will have terms of the form [math] 1/x, 1/x^2, 1/(x+1),1/(x+1)^2,1/(x+1)^3 [/math] - out of these, [math]1/x, 1/(x+1) [/math] will give irrational integrands. Thus, these cannot appear in your partial fraction expansion. Therefore [math]\frac{f(x)}{x^2 (1+x)^3} = \frac{a}{x^2}+\frac{b}{(x+1)^2}+[/math] [math] \frac{c}{(x+1)^3} [/math][math] \implies \ f(x) = a(x+1)^3+b x^2(x+1)+c x^2 [/math]Imposing the condition that [math] f(x) [/math] is quadratic and its value at 0 is 1, we obtain[math] a+b = 0, a = 1, b =- 1 [/math]Now, you can easily evaluate [math]f'(x) = 3a(x+1)^2+b x^2+2b x(x+1)+2cx^2 [/math] and thus [math] f'(0) = 3a = 3 [/math].