Which of the following describes the graphs of 2x+5y=9 and 10x=4y+18?
2x+5y=9 y=-(2/5)x+9/5-----(1) Gradient on (1) m1= -(2/5) 10x=4y+18 y=(5/2)x-9/2----(2) Gradient on (2) m2= (5/2) m1*m2= -(2/5)*(5/2)=-1 so...graphs are perpendicular
I need help on this ? Use the following quadratic equation to complete parts a – c below. ?
a) 2x^2 + 10x +12 = 0 2( x^2 +5x+6) = 0 x^2 + 2x + 3x+6 = 0 x(x+2) +3(x+2) = 0 (x+2) (x+3) = 0 [ thats factorised] so x = -2 or x = -3 b) now using the quadratic formula, x = (-10 +/- √10^2 - 4 x 12 x 2) / 4 = (-10 +/- √100 -96) / 4 = (-10 +/- √4) / 4 taking +ve x = (-10 + 2)/4 = -2 taking -ve x = (-10-2) /4 = -3 c) the realation is probably that the answers are the same hope that helped;)
What are the solutions to the equation 10x^8-1=0?
10x^8 - 1 = 010x^8 = 1x^8 = 1/10x = (1/10)^1/8x = 1/1.33352143216x = 0.74989420933
What are the solutions to the equation 10x^7-1=0?
10x^7 - 1 = 0add +1 both sides to eliminate -1.Now it will be 10x^7 = 1divide by 10 both sides to get x by itself.x^7 = 1/10x = 7*sqrt(1/10)answer is 7*sqrt(1/10)
For quadratic equation solutions, how do I find if one answer doesn't work?
10x² -25y²- 100x +160 = 0 y² -2x + 16 = 0 y² = 2x - 16 10x² -25(2x - 16)- 100x +160 = 0 10x² -50x + 400 - 100x +160 = 0 10x² -50x - 100x + 400 +160 = 0 10x² - 150x + 560 = 0 10(x² - 15x + 56) = 0 10(x-7)(x-8) = 0 x = 7, 8 Check: Given ax² + bx + c = 0, quadratic formula: x = [-b ± √(b²-4ac)]/2a a = 10, b = -150, c = 560 x = [-(-150) ± √((-150)²-4(10)(560))]/2(10) x = [150 ± √(22500-22400)]/20 x = [150 ± √(100)]/20 x = [150 ± 10]/20 x = [15 ± 1]/2 x = 16/2 or 14/2 x = 8 or 7 When x = 7: y² = 2x - 16 y² = 2(7) - 16 y² = 14 - 16 y² = -2 y = i√2 When x = 8: y² = 2x - 16 y² = 2(8) - 16 y² = 16 - 16 y² = 0 y = 0 Answers: (8, 0) and (7, i√2) =========================== Check for (8, 0): 10(8)² - 25(0)² - 100(8) +160 = 0 10*64 - 0 - 800 +160 = 0 640 - 0 - 800 +160 = 0 0 = 0 True y² -2x + 16 = 0 (0) - 2(8) + 16 = 0 0 - 16 + 16 = 0 0 = 0 True ================== Check for (7, i√2): 10(7)² - 25(i√2)² - 100(7) +160 = 0 10*49 - 25(-2) - 700 +160 = 0 490 + 50 - 700 +160 = 0 0 = 0 True y² -2x + 16 = 0 (i√2)² -2(7) + 16 = 0 (-2) - 14 + 16 = 0 -2 - 14 + 16 = 0 -16 + 16 = 0 0 = 0 True
How do you solve the equation 10x^9-16x^5-10x=0?
x ( 10x^8 - 16x^4 - 10) =0Let x^4=tx2(5t^2–8t -5)=0t1 =8/10+(64+100)^1/2)/10t1 =0.8+ 1.28= 2.08t2 =0.8- 1.28= - 0.48x1 =0x2= +(2.08)^4=+1.2x3= - 1.2
How do you solve the equation 16x^12-17x^18-19 =0?
I am going to assume that this question was mistakenly typed. With the [math]-19[/math] and no [math]-19x^6,[/math] there is no solution that I can think of as general polynomials have no closed form solution after power [math]x^5[/math]. If this is typed correctly, then you will need graduate level real analysis to solve this…Assuming that it is suppose to be[math]-19x^6[/math]:All powers have a factor of 6, so factor [math]x^6[/math] out.You are left with [math]-17x^3 + 16x^2 - 19 = 0.[/math] Solve the cubic equation for the roots.Your solutions are x = 0 and the cubic roots.
Complete the square on the following quadratic [math]x^2+10x+8=0[/math] What value is added to both sides of the equation?
17A2AWhat is half of the “b value” then squared? The B value is 10, from 10x in the middle. Half of 10 is 5. 5 squared is 25. Right now you have 8 but you want 25, so 25 -8 = 17.x^2 + 10x +8 = 0x^2 + 10x + 8 + 17 = 0 + 17x^2 +10x + 25 = 17(x + 5)^2 = 17
For what of K, the system of equations, 5x-ky=3 and 10x-3y=6 is independent?
The second equation can be rearranged to give y = 10x/3 - 2 and the slope of the line m2 = 10/3. Rewriting the first equation we obtain y = 5x/k - 3/k ad its slope m1 = 5/k. For the two equations to be independent their slopes cannot be equal. There must be one and only one solution for the two equations where they intersect. That is, m1 cannot equal m2. Substituting their values, 5/k cannot equal 10/3. Which means k cannot equal 3/2. In fact if k = 3/2 the two equations are identical. Any other value for k makes the slopes different and the two lines will intersect at some point.
How many common roots are their between these two equation [math]x^3 + 4x^2 + 5x + 18 = 0[/math] and [math]x^3 + 3x^2 + 10x + 12 = 0[/math], and what are the roots?
Zero.Let's assume both equations have a common root = r.Since r is a root for both equations, it would individually satisfy both equations. So,r^3 + 4r^2 + 5r + 18 = 0 --- (eq 1)r^3 + 3r^2 + 10r + 12 = 0 --- (eq 2)Clearly, eq1 = eq2 = 0So,r^3 + 4r^2 + 5r + 18 = r^3 + 3r^2 + 10r + 12which leads to a quadratic equationr^2 -5r + 6 = 0Its solutions are 2, 3But you will very easily see conflict, none of these solution (2 or 3) satisfies eq1 or eq2. So, our very first assumption of having a common root was wrong.