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77julia77 [94]
3 years ago
10

20=6x+14 answer 0=6 true or false ?????

Mathematics
2 answers:
Roman55 [17]3 years ago
4 0

Hey there!

Let's solve the equation.

20=6x+14

Subtract 14 from both sides.

6=6x

Divide 6 on both sides.

x=1

This is false.

I hope this helps!

~kaikers

Bogdan [553]3 years ago
4 0

[|] Answer [|]

\boxed{False}

[|] Explanation [|]

20 = 6x + 14

0 = 6

True or False

_________________

_________________

- Solve For X -

Switch Sides:

6x + 14 = 20

Subtract 14 From Both Sides:

6x + 14 - 14 = 20 - 14

Simplify:

6x = 6

Divide Both Sides By 6:

\frac{6x}{6} \ = \frac{6}{6}

Simplify:

X = 1

__________________

__________________

X \boxed{Is \ **NOT**} Equal To 6

\boxed{[|] \ Eclipsed \ [|]}

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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
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