A)
b)
since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP
thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)
c)
what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?
Answer:
Hope i helped :)
Step-by-step explanation:
8) T-V =24
24 -6 =18
2x-18 =18
2x=0
x= 0/2
x=0
9)
*LMK
*L-K=x+3
*L-M=6x-10
*M-K= 3x-3
L-M + M-K = 6x -10 + 3x -3 = 9x -13
10 and 11)
Both lines are horizontal
both lines consist of intervals due to having 2 points on one line making it a distance between 2 points.
they are also concurrent lines because more then 3 lines pass through one point
they are also intersecting lines because they cross over each other
I think it might be grow. The slope is a positive number.
Answer:
(m − 3) (m + 2)
Step-by-step explanation:
m² − m − 6
To factor a quadratic ax² + bx + c, you can use the AC method.
1. Multiply a and c.
2. Find factors of ac that add up to b.
3. Divide the factors by a and reduce.
4. The denominators are the coefficients, the numerators are the constants.
Here, a = 1, b = -1, and c = -6.
1. ac = -6
2. Factors of -6 that add up to -1 are -3 and 2.
3. -3/1, 2/1
4. Factors are m − 3 and m + 2.
m² − m − 6 = (m − 3) (m + 2)
Answer:
sin θ/2=5√26/26=0.196
Step-by-step explanation:
θ ∈(π,3π/2)
such that
θ/2 ∈(π/2,3π/4)
As a result,
0<sin θ/2<1, and
-1<cos θ/2<0
tan θ/2=sin θ/2/cos θ/2
such that
tan θ/2<0
Let
t=tan θ/2
t<0
By the double angle identity for tangents
2 tan θ/2/1-(tanθ /2)^2 = tanθ
2t/1-t^2=5/12
24t=5 - 5t^2
Solve this quadratic equation for t :
t1=1/5 and
t2= -5
Discard t1 because t is not smaller than 0
Let s= sin θ/2
0<s<1.
By the definition of tangents.
tan θ/2= sin θ/2/ cos θ/2
Apply the Pythagorean Algorithm to express the cosine of θ/2 in terms of s. Note the cos θ/2 is expected to be smaller than zero.
cos θ/2 = -√1-(sin θ/2)^2 = - √1-s^2
Solve for s.
s/-√1-s^2 = -5
s^2=25(1-s^2)
s=√25/26 = 5√26/26
Therefore
sin θ/2=5√26/26=0.196....
