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Blizzard [7]
3 years ago
12

Given six consecutive integers with a sum of seven times the

Mathematics
1 answer:
Furkat [3]3 years ago
6 0

Answer:

<em>x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 7(x + 1)</em>

Step-by-step explanation:

<u>Equations</u>

The situation can be written as an algebraic equation by setting the variables as follows:

x = first integer

x + 1 = second integer

x + 2 = third integer

x + 3 = fourth integer

x + 4 = fifth integer

x + 5 = sixth (and last) integer

The sum of all six numbers must be equal to 7 times the second number, thus:

x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 7(x + 1)

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Please help :)))) ( attachment )
Anastaziya [24]
Let,
f(x) = -2x+34
g(x) = (-x/3) - 10
h(x) = -|3x|
k(x) = (x-2)^2

This is a trial and error type of problem (aka "guess and check"). There are 24 combinations to try out for each problem, so it might take a while. It turns out that 

g(h(k(f(15)))) = -6
f(k(g(h(8)))) = 2

So the order for part A should be: f, k, h, g
The order for part B should be: h, g, k f
note how I'm working from the right and moving left (working inside and moving out).


Here's proof of both claims

-----------------------------------------

Proof of Claim 1:

f(x) = -2x+34
f(15) = -2(15)+34
f(15) = 4
-----------------
k(x) = (x-2)^2
k(f(15)) = (f(15)-2)^2
k(f(15)) = (4-2)^2
k(f(15)) = 4
-----------------
h(x) = -|3x|
h(k(f(15))) = -|3*k(f(15))|
h(k(f(15))) = -|3*4|
h(k(f(15))) = -12
-----------------
g(x) = (-x/3) - 10
g(h(k(f(15))) ) = (-h(k(f(15))) /3) - 10
g(h(k(f(15))) ) = (-(-12) /3) - 10
g(h(k(f(15))) ) = -6

-----------------------------------------

Proof of Claim 2:

h(x) = -|3x|
h(8) = -|3*8|
h(8) = -24
---------------
g(x) = (-x/3) - 10
g(h(8)) = (-h(8)/3) - 10
g(h(8)) = (-(-24)/3) - 10
g(h(8)) = -2
---------------
k(x) = (x-2)^2
k(g(h(8))) = (g(h(8))-2)^2
k(g(h(8))) = (-2-2)^2
k(g(h(8))) = 16
---------------
f(x) = -2x+34
f(k(g(h(8))) ) = -2*(k(g(h(8))) )+34
f(k(g(h(8))) ) = -2*(16)+34
f(k(g(h(8))) ) = 2
5 0
4 years ago
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