Answer:
5 km.
Step-by-step explanation:
We have to calculate the distance if we apply the speed formula that would be the distance in a given time:
v = d / t
therefore, if we solve for distance it would be:
d = v * t
Now the speed is 6 km / h but the time must be calculated, it leaves at 9:15 and returns at 10:05, that is, there is no exact time but 50 minutes, one hour equals 60 minutes, therefore:
50 min * 1 h / 60 min = 0.833 h, that is, the child took 0.833 hours to get to school, now if we replace:
d = 6 * 0.8333
d = 5
In other words, the distance between the school and the house is 5 km.
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.
