Answer:
1) £2 = €2.32
£5 = €5.80
£50 = €58
2) The graph will be a straight line
3) (0, 0)
4) Label the independent variable, £ on the x-axis and dependent variable € on the y-axis
Step-by-step explanation:
1) The given conversion factors is £1 = €1.16
Therefore;
£2 = 2 × €1.16 = €2.32
£2 = €2.32
£5 = 5 × €1.16 = €5.80
£5 = €5.80
£50 = 50 × €1.16 = €58
£50 = €58
2) The shape of the plot of the directly proportional currencies graph will be a straight line
3) Given that the £ is directly proportional to the € and that the value of the € can be found directly by multiplying the amount in £ by 1.16, without the addition of a constant, the graph crosses the axes at the origin (0, 0)
4) The y-axes which is the dependent variable should be labelled €, while the x-axis which is the independent variable should be labelled £
The slope and speed of Plane A is 375mph and slope and speed for Plane B is 425 mph... so Plane A travels the fastest
X➝-x by reflection, so g(x)=-x
This problem can be readily solved if we are familiar with the point-slope form of straight lines:
y-y0=m(x-x0) ...................................(1)
where
m=slope of line
(x0,y0) is a point through which the line passes.
We know that the line passes through A(3,-6), B(1,2)
All options have a slope of -4, so that should not be a problem. In fact, if we check the slope=(yb-ya)/(xb-xa), we do find that the slope m=-4.
So we can check which line passes through which point:
a. y+6=-4(x-3)
Rearrange to the form of equation (1) above,
y-(-6)=-4(x-3) means that line passes through A(3,-6) => ok
b. y-1=-4(x-2) means line passes through (2,1), which is neither A nor B
****** this equation is not the line passing through A & B *****
c. y=-4x+6 subtract 2 from both sides (to make the y-coordinate 2)
y-2 = -4x+4, rearrange
y-2 = -4(x-1)
which means that it passes through B(1,2), so ok
d. y-2=-4(x-1)
this is the same as the previous equation, so it passes through B(1,2),
this equation is ok.
Answer: the equation y-1=-4(x-2) does NOT pass through both A and B.
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
_______________
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
______________
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
________________
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.
