They are: 1, 2, 4, 8, a, 2a, 4a and 8a.
1. The y-intercept of a parabola with equation y = ax^2 + bx + c is c. So, c = -22. The vertex is (x, y) = (-4, 2), where x = - b/2ax = - b/2a-4 = - b/2a(-4)(-2a) = (-b/2a)(-2a)8a = bSo we have:y = ax^2 + bx + c (substitute what you know: 2 for y, -4 for x, 8a for b, and -2 for c)2 = a(-4)^2 + (8a)(-4) + (-2)2 = 16a - 32a - 22 = - 16a - 2 (add 2 to both sides)4 = -16a (divide by -16 to both sides)-1/4 = aSince b = 8a, then b = 8(-1/4) = -2 Since a = -1/4, b = -2, and c = -2, then we can write the equation of the parabola asy = (-1/4)x^2 - 2x - 2.
-4 + 10 + 6a - 2a + 3 9 + 4a
7x - 8a = 3x + 24a First, add 8a to both sides: 7x = 3x + 24a + 8a Now subtract 3x from both sides to get all x terms on one side: 7x - 3x = 24a + 8a Simplify... 4x = 32a And divide both sides by 4: x = 32a / 4 = 8a x = 8a
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
They are: 1, 2, 4, 8, a, 2a, 4a and 8a.
If 8a = 40 then a must = 5So: 2*5-4 = 6
Suppose the middle integer is 2a. Then the smallest is 2a-2 and the biggest is 2a+2. 4 times the smallest is 8a-8 So largest subtracted from the smallest is (8a-8) - (2a+2) = 6a-10 So, 6a-10 = 2*2a = 4a so that 2a = 10 So the integers are 8, 10 and 12.
-4 + 2A = 12 - 15A + A-4 + 2A = 12 - 16A2A = 12 - 16A + 42A = 16 - 16AA = 8 - 8A
As all the required numbers are even, we can identify the smallest value number as 2a, then the other three consecutive even numbers are (2a + 2), (2a + 4) and (2a + 6). The sum of these four numbers is, 2a + (2a + 2) + (2a + 4) + (2a + 6) = 8a + 12 Then, 8a + 12 = 412 8a = 412 - 12 = 400 2a = 400 / 4 = 100 The four consecutive even numbers are 100, 102, 104 & 106. NOTE : In this problem the choice of 'a' as the smallest value number would have achieved the correct answer but in other situations such a choice could have resulted in 4 consecutive ODD numbers being found. When in doubt use 2a or similar to clearly identify that the number is even.
32a + 8= 8(4a+1)
1. The y-intercept of a parabola with equation y = ax^2 + bx + c is c. So, c = -22. The vertex is (x, y) = (-4, 2), where x = - b/2ax = - b/2a-4 = - b/2a(-4)(-2a) = (-b/2a)(-2a)8a = bSo we have:y = ax^2 + bx + c (substitute what you know: 2 for y, -4 for x, 8a for b, and -2 for c)2 = a(-4)^2 + (8a)(-4) + (-2)2 = 16a - 32a - 22 = - 16a - 2 (add 2 to both sides)4 = -16a (divide by -16 to both sides)-1/4 = aSince b = 8a, then b = 8(-1/4) = -2 Since a = -1/4, b = -2, and c = -2, then we can write the equation of the parabola asy = (-1/4)x^2 - 2x - 2.
4(2a+b) + 2(6a+2b) Working this out it is: 8a + 4b + 12a + 4b = 20a + 8b or 4(5a+2b).
-4 + 10 + 6a - 2a + 3 9 + 4a
The area, in terms of a, is 2a*(3a-4) = 6a2-8a However, it is not possible to give its area in terms of x without knowing anything about x, such as how x relates to a.
2a-b equals 17 3a plus 4b equals -13? Elimination Method If you can multiply one of the equations by a number and eliminate one of the variables, that is an easy to find the value of the other variable. I see -b in Eq. #1 and +4b in Eq. #2. I can multiply by Eq. #1 by 4 and add to Eq. #2 eliminate b. Eq. #1 = 2a - b = 17 Eq. #2 = 3a + 4b = -13 4* (2a - b = 17) = 8a - 4b = 68 Now add 4* Eq. #1 to Eq. #2 ..8a - 4b = 68 +.3a + 4b = -13 .11a + 0b = 55 11a = 55 divide both sides by 11 a = 5 Substituting a = 5 into Eq. #1 (2 * 5) - b = 17 10 - b = 17 subtract 10 from both sides - b = 17 - 10 -b = 7 Divide by -1 b = -7 Check in Eq. #2, by substituting a = 5 and b = -7 3a + 4b = -13 (3 * 5) + (4 * -7) = -13 15 + -28 = -13 -13 = -13 The other way of solving simultaneous equations is Substitution. Solve for one variable in terms of the other variable. Eq. #1 = 2a - b = 17 Eq. #2 = 3a + 4b = -13 Solve Eq.#1 for a in terms of b. 2a - b = 17 add +b to both sides 2a = b + 17 divide (b + 17) by 2 a = (b + 17) ÷ 2 Substitute a = (b + 17) ÷ 2 into Eq.#2, and solve for b Eq. #2 = 3a + 4b = -13 3 * [(b + 17) ÷ 2] + 4b = -13 Multiply 3 * (b + 17) (3b + 51)÷2 + 4b = -13 Multiply both sides by 2 to eliminate the (÷2) (3b + 51) + 8b = -26 Add (3b + 8b) 11b + 51 = -26 Add -51 to both sides 11b = -77 Divide both sides by 11 b = -7 Substitute b = -7 into Eq.#1 Eq. #1 = 2a - b = 17 2a + 7 = 17 Subtract 7 from both sides 2a = 10 Divide both sides by 2 a = 5 That Substitution sure was complicated compared to the Elimination method. I wonder if it would have been easier to solve Eq.#1 for a in terms of b. That [(b + 17) ÷ 2] was the trouble maker. Eq. #1 = 2a - b = 17 Add b to both sides 2a = 17 + b subtract 17 from both sides 2a -17 = b reverse sides b = 2a - 17 Substitute (b = 2a - 17) into Eq.#2, and solve for a Eq. #2 = 3a + 4b = -13 3a + 4 * (2a - 17) = -13 Multiply 4 * (2a - 17) 3a + (8a - 68) = -13 Add 3a + 8a 11a - 68 = -13 Add +68 to both sides 11a = 55 Divide by 11 a = 5 Eq. #1 = 2a - b = 17 Substitute a = 5 into Eq.#1 2 * 5 - b = 17 Multiply 2 * 5 = 10 10 - b = 17 Add -10 to both sides b = -7 We got the same answers as we did by using the elimination method, but the substitution was more work. Sometimes using the elimination method, you have to multiply both equations by 2 different numbers to eliminate a variable, but I still think it is easier.
A = 4