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xxTIMURxx [149]
3 years ago
13

How can I simplify/rewrite the 1st(top) equation into the 2nd(Botton),please show a step by step solution.​

Mathematics
1 answer:
Vaselesa [24]3 years ago
4 0

Answer:

see below

Step-by-step explanation:

-2(4 - \frac{x}{2} )^{5} + C

Think of the above as:

  -\frac{2}{1} * (\frac{8}{2} - \frac{x}{2} )^{5} + C

= -\frac{2}{1} * \frac{(8 - x)^{5}}{2^{5}}  + C

= -\frac{2}{1} * \frac{(8 - x)^{5}}{32}  + C

= -\frac{1}{1} * \frac{(8 - x)^{5}}{16}  + C

=   \frac{-1(8 - x)^{5}}{16}  + C

=   \frac{(-8 + x)^{5}}{16}  + C

Can flip it around to be...

\frac{(x - 8)^{5}}{16}  + C

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L = J + 6
L = 15 + 6
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So Leah spent $21

5 0
3 years ago
The difference between the roots of the quadratic equation x2 -14x +q =0 is 6. Find q
Irina-Kira [14]

9514 1404 393

Answer:

  q = 40

Step-by-step explanation:

When the quadratic has roots p and r, it can be factored as ...

  (x -p)(x -r) = x² -(p+r)x +pr

So, the sum of the roots is 14, and their difference is 6. This lets us find the roots from ...

  p + r = 14

  p - r = 6

  2p = 20 . . . add the two equations

  p = 10

  r = 14 -p = 4

The value of interest is then ...

  q = pr = (10)(4)

  q = 40

__

The graph shows the roots to be 4 and 10, as we found.

3 0
3 years ago
Solve for (f+g)(5) if f(x)=6x+3 and g(x)=x-4 show you work please
Diano4ka-milaya [45]
There are two ways to do this.

The first way is to algebraically find (f+g)(x) first and plug in x = 5 later. Doing that method leads us to
(f+g)(x) = f(x) + g(x)
(f+g)(x) = 6x+3 + x-4
(f+g)(x) = 7x-1
(f+g)(5) = 7(5)-1
(f+g)(5) = 34

OR

you can compute f(5) and g(5) first, then add up those sub-results to get
f(x) = 6x+3
f(5) = 6(5)+3
f(5) = 33
g(x) = x-4
g(5) = 5-4
g(5) = 1
Adding up these results gives: (f+g)(5) = f(5) + g(5) = 33+1 = 34

Either way, the final answer is 34

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Step-by-step explanation:

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