Q: How do you solve 9a plus 4a equals 26?

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The general equation of a parabola is y = ax2 + bx + c.The vertex of a parabola is (-b/2a, c - b2/4a). In our case the vertex is (2, -4). So we have,-b/2a = 2, so that b = -4a, andc - b2/4a = -4 (substitute -4a for b)c - (-4a)2/4a = -4c = 16a2/4a = -4 (simplify)c - 4a = -4 (solve for c)c = 4a - 4By substituting -4a for b, and 4a - 4 for c, the equation becomes,y = ax2 + bx + cy = ax2 + (-4a)x + 4a - 4Since it is given that when x = -3, y = -3, we substitute them into the new equation and solve it for a. So we have,y = ax2 + (-4a)x + 4a - 4-3 = a(-3)2 + (-4a)(-3) + 4a - 4-3 = 9a + 12a + 4a - 4 (add 4 to both side, and add alike terms on the right-hand side)1 = 25a (divide by 25 to both sides)1/25 = a (this is the required answer)If you want to find the equation of the parabola, just substitute 1/25 for a, such as:y = ax2 + (-4a)x + 4a - 4y = (1/25)x2 + [-4(1/25)]x + 4(1/25) - 4y = (1/25)x2 -4/25x + 4/25 - 4y = (1/25)x2 -4/25x + 96/25

Simultaneous Equations 4a - 5b = 7 4a + 5b = 17 Add the two , this will eliminate 'b' 8a = 24 Divide both sides by '8' a = 3 When a= 3 , substitute into eityher equation for 'b'. 4(3) + 5b = 17 12 + 5b = 17 5b = 17 - 12 = 5 5b = Divide both sides by '5'. Hence 'b = 1'.

Let us say that f(x)=x^4A derivative is the opposite to an integral.If you were to integrate x^4, the first process is taking the power [which in this case is 4], multiplying it by any value before the x [which is 1], then subtracting 1 from the initial power [4]. This leaves 4x^3. The final step is taking the integral of what is 'inside' the power [which is (x)], and multiplying this to the entire answer, which results in 4x^3 x 1 = 4x^3If you were to derive (x)^4, you would just add 1 to the power [4] to become (x)^5 then put the value of the power as the denominator and the function as a numerator. This leaves [(x^5)/(5)]To assure that the derivative is correct, integrate it. (x^5) would become 5x^4. Since (x^5) is over (5), [(5x^4)/(5)] cancels the 5 on the numerator and denominator, thus leaving the original function of x^4

Related questions

Solve the following equation for A : 2A/3 = 8 + 4A

4a + a = 25 add 4a and a 5a = 25 then divide 25 by 5 a = 5

9=4a+13 9-13=4a -4=4a -4/4=a -1=a a=-1

11a+9 = 4a+30 11a-4a = 30-9 7a = 21 Solution: a = 3

4a + 2 = 4(6) + 2 = 24 + 2 = 26

4a - 2.1 = 9a + 5.4 4a - 9a = 5.4 +2.1 -5a = 7.5 -a = 1.5 a = -1.5

8a+4a=144 12a=144 12a/12=144/12 A=144

4a - 9 = 27 4a = 36 a = 9

Add the equations: 4a + 4a - 5b + 5b = 7 + 17 ie 8a = 24 a = 3, so b = 1

a + 4 = 10(a + 4) - 4 = 10 - 4a = 6

4a - 5b = 74a +5b = 17From the first equation:4a = 7 + 5bsubstitute this value in the second equation:7 + 5b + 5b = 177 + 10b = 1710b = 10b = 1Now,solve for a :4a - 5(1) = 74a = 7 + 54a = 12a = 3

3a + 4 = 4a - 3 4 = a -3 (subtract 3a from each side) 7 = a (add 3 to both sides) a = 7