elsauceomotho
02.01.2020 •
Mathematics
What are the solutions for the equation 2z^2+4z+2=0
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Ответ:
2(z^2+2z+1)=0
factor
2(z+1)(z+1)=0
se to zero
2=0 false
z+1=0
z=-1
solution is z=-1
Ответ:
a) g(x) = 2x-2
b) g(x) = 8x-24
c) g(x) =2x-10
d) g(x) = 8x-6
Step-by-step explanation:
Given Shift f(x) four units right
Given function f(x) = 2x-6
a) The function f(x) shifts '4' units right
g(x) = 2x - 6+4 = 2x -2
b) f(x) by a factor of 1/4 towards the y-axis
Given function f(x) = 2x -6
The new function
g(x) = 4(2x-6) = 8x-24
c)
Given shifts f(x) four units down
g(x) = 2x-6-4 = 2x -10
d) Stretches by a factor '4'
f(4x) = 4(2x-6) = 8x -6
g(x) =8x -6