Here is my work! the answer is (-3x+5)(2x-1)
20 units rotated to 90 cc reflected over the x axis and translated 6 units down is ?
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Answer:
We know that for a line:
y = a*x + b
where a is the slope and b is the y-intercept.
Any line with a slope equal to -(1/a) will be perpendicular to the one above.
So here we start with the line:
3x + 4y + 5 = 0
let's rewrite this as:
4y = -3x - 5
y = -(3/4)*x - (5/4)
So a line perpendicular to this one, has a slope equal to:
- (-4/3) = (4/3)
So the perpendicular line will be something like:
y = (4/3)*x + c
We know that this line passes through the point (a, 3)
this means that, when x = a, y must be equal to 3.
Replacing these in the above line equation, we get:
3 = (4/3)*a + c
c = 3 - (4/3)*a
Then the equation for our line is:
y = (4/3)*x + 3 - (4/3)*a
We can rewrite this as:
y = (4/3)*(x -a) + 3
now we need to find the point where this line ( y = -(3/4)*x - (5/4)) and the original line intersect.
We can find this by solving:
(4/3)*(x -a) + 3 = y = -(3/4)*x - (5/4)
(4/3)*(x -a) + 3 = -(3/4)*x - (5/4)
(4/3)*x - (3/4)*x = -(4/3)*a - 3 - (5/4)
(16/12)*x - (9/12)*x = -(4/3)*a - 12/4 - 5/4
(7/12)*x = -(4/13)*a - 17/4
x = (-(4/13)*a - 17/4)*(12/7) = - (48/91)*a - 51/7
And the y-value is given by inputin this in any of the two lines, for example with the first one we get:
y = -(3/4)*(- (48/91)*a - 51/7) - (5/4)
= (36/91)*a + (153/28) - 5/4
Then the intersection point is:
( - (48/91)*a - 51/7, (36/91)*a + (153/28) - 5/4)
And we want that the distance between this point, and our original point (3, a) to be equal to 4.
Remember that the distance between two points (a, b) and (c, d) is:
distance = √( (a - c)^2 + (b - d)^2)
So here, the distance between (a, 3) and ( - (48/91)*a - 51/7, (36/91)*a + (153/28) - 5/4) is 4
4 = √( (a + (48/91)*a + 51/7)^2 + (3 - (36/91)*a + (153/28) - 5/4 )^2)
If we square both sides, we get:
4^2 = 16 = (a + (48/91)*a + 51/7)^2 + (3 - (36/91)*a - (153/28) + 5/4 )^2)
Now we need to solve this for a.
16 = (a*(1 + 48/91) + 51/7)^2 + ( -(36/91)*a + 3 - 5/4 + (153/28) )^2
16 = ( a*(139/91) + 51/7)^2 + ( -(36/91)*a - (43/28) )^2
16 = a^2*(139/91)^2 + 2*a*(139/91)*51/7 + (51/7)^2 + a^2*(36/91)^2 + 2*(36/91)*a*(43/28) + (43/28)^2
16 = a^2*( (139/91)^2 + (36/91)^2) + a*( 2*(139/91)*51/7 + 2*(36/91)*(43/28)) + (51/7)^2 + (43/28)^2
At this point we can see that this is really messy, so let's start solving these fractions.
16 = (2.49)*a^2 + a*(23.47) + 55.44
0 = (2.49)*a^2 + a*(23.47) + 55.44 - 16
0 = (2.49)*a^2 + a*(23.47) + 39.44
Now we can use the Bhaskara's formula for quadratic equations, the two solutions will be:
Then the two possible values of a are:
a = (-23.47 + 12.57)/4.98 = -2.19
a = (-23.47 - 12.57)/4.98 = -7.23
Answer:
0.8041
Step-by-step explanation:
We know that
μ=72 and σ=5
and P(65<X<78)
We can determine the Z value as (X-μ)/σ
P( 65<X<78 )=P( 65-72< X-μ<78-72)
To fine the Z values:
From the standard normal tables:
to find P ( Z<-1.4)
From the standard normal tables:
Therefore