Answer:
Step-by-step explanation:
3ab + 7a² + 2b² - (7a² - 4ab + 2b²) = 3ab + 7a² + 2b² - 7a² + 4ab - 2b²
Now combine like term. Like terms have same variable with same power.
= <u>3ab + 4ab</u> <u>+ 7a² - 7a²</u> + <u>2b² - 2b²</u>
= 7ab
rewrite the equation set = to 0.
x^2 + 5x - 8 = 0
The quadratic will not factor so you have to use the quadratic formula.
x = (-b + - sqrt(b^2 - 4ac))/2a
x = (5 + - sqrt(25 - 4* 1* -8))/2
x = (5 + - sqrt 57)/2
The x2is not the same as 2x. It is x^2. X tot he second power which makes the problem a quadratic equation. You cannot combine the terms x^2 and -5x because they so not have the same power.
<span> To get an explicit formula, we need to find an expression which gives the n-th term without having to compute earlier terms in the sequence. Looking at the numbers, and from the recursive formula, we see that the sequence is built by subtracting n from the previous term. This is similar to the triangular number sequence 1,3,6,10,15,... which has the explicit formula a_n = n(n+1)/2. In our case we are subtracting n from the previous term, so we multiply by -1/2 instead of 1/2. However, we also need to add a constant term to reproduce the numbers of the sequence. We can write a_1 = -1(2)/2 + c = 8. Therefore, c = 9.
So the explict formula is:
a_n = -n(n+1)/2 + 9</span>
Answer:
0.46666666666
Step-by-step explanation:
i tried
Answer:
- hits the ground at x = -0.732, and x = 2.732
- only the positive solution is reasonable
Step-by-step explanation:
The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.
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<h3>graphing</h3>
The attachment shows a graphing calculator solution to the equation
-3x^2 + 6x + 6 = 0
The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.
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<h3>completing the square</h3>
We can also solve the equation algebraically. One of the simplest methods is "completing the square."
-3x^2 +6x +6 = 0
x^2 -2x = 2 . . . . . . . . divide by -3 and add 2
x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square
(x -1)^2 = 3 . . . . . . . . written as a square
x -1 = ±√3 . . . . . . . take the square root
x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground
The numerical values of these solutions are approximately ...
x ≈ {-0.732, 2.732}
The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.