Answer:
Step-by-step explanation:
First, you need to combine like terms in order to begin simplifying. This is done by subtracting 2x from 6x. Also, add 7 to 23.
6x + 10 - 2x = 7 + 23
4x + 10 = 7 + 23
4x + 10 = 30
Now, you have a two-step equation. Subtract 10 from both sides, and then divide by 4.
4x + 10 = 30
4x = 20
x = 5 is the final answer.
Let's to the first example:
f(x) = x^2 + 9x + 20
Ussing the formula of basckara
a = 1
b = 9
c = 20
Delta = b^2 - 4ac
Delta = 9^2 - 4.(1).(20)
Delta = 81 - 80
Delta = 1
x = [ -b +/- √(Delta) ]/2a
Replacing the data:
x = [ -9 +/- √1 ]/2
x' = (-9 -1)/2 <=> - 5
Or
x" = (-9+1)/2 <=> - 4
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Already the second example:
f(x) = x^2 -4x -60
Ussing the formula of basckara again
a = 1
b = -4
c = -60
Delta = b^2 -4ac
Delta = (-4)^2 -4.(1).(-60)
Delta = 16 + 240
Delta = 256
Then, following:
x = [ -b +/- √(Delta)]/2a
Replacing the information
x = [ -(-4) +/- √256 ]/2
x = [ 4 +/- 16]/2
x' = (4-16)/2 <=> -6
Or
x" = (4+16)/2 <=> 10
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Now we are going to the 3 example
x^2 + 24 = 14x
Isolating 14x , but changing the sinal positive to negative
x^2 - 14x + 24 = 0
Now we can to apply the formula of basckara
a = 1
b = -14
c = 24
Delta = b^2 -4ac
Delta = (-14)^2 -4.(1).(24)
Delta = 196 - 96
Delta = 100
Then we stayed with:
x = [ -b +/- √Delta ]/2a
x = [ -(-14) +/- √100 ]/2
We wiil have two possibilities
x' = ( 14 -10)/2 <=> 2
Or
x" = (14 +10)/2 <=> 12
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To the last example will be the same thing.
f(x) = x^2 - x -72
a = 1
b = -1
c = -72
Delta = b^2 -4ac
Delta = (-1)^2 -4(1).(-72)
Delta = 1 + 288
Delta = 289
Then we are going to stay:
x = [ -b +/- √Delta]/2a
x = [ -(-1) +/- √289]/2
x = ( 1 +/- 17)/2
We will have two roots
That's :
x = (1 - 17)/2 <=> -8
Or
x = (1+17)/2 <=> 9
Well, this would be your answers.
Answer:
you got this Playa I believe in you
Answer:
or 
Step-by-step explanation:
By definition the function
has a domain of 
.
Since the function is not defined for the values of x negative or equal to zero.
The range of this function is all the real numbers.
Since 
when 
and 
when 
In this case we have the function 
Therefore since the log function is squared then its range is now
Answer:
arithmitic
Step-by-step explanation:
