Given that,
KM bisects ∠LKJ.
m∠LKM = 4x + 12 and m∠MKJ = 6x - 6
To find,
The value of x.
Solution,
As KM bisects ∠LKJ. It would mean that,
m∠LKM = m∠MKJ
Putting the above values,
4x+12=6x-6
We can solve the above equations as follows :
4x-6x=-12-6
-2x=-18
x=9
Therefore, the value of x is equal to 9.
Answer:
The value of f(2) is 1
Step-by-step explanation:
Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)
n+1 </span>≡ 1 ( mod 3)...(2)
n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)
n+1 ≡ 2 ( mod 3)...(2)
n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)
n+1 ≡ 3 ( mod 3)...(2)
n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)
Answer:
G. ABD = 74
H. DBC = 206
I. XYW = 33.75
J. WYZ = 46.25
Step-by-step explanation:
For G and H: You have a straight line (ABC) with another line coming off of it, creating two angles (ABD and DBC). A straight line has an angle of 180 degrees. This means that the two angles from the straight line when combined will give you 180 degrees. Solve for x.
ABD + DBC = ABC
(1/2x + 20) + (2x - 10) = 180
1/2x + 20 + 2x - 10 = 180
5/2x + 10 = 180
5/2x = 170
x = 108
Now that you have x, you can solve for each angle.
ABD = 1/2x + 20
ABD = 1/2(108) + 20
ABD = 54 + 20
ABD = 74
DBC = 2x - 10
DBC = 2(108) - 10
DBC = 216 - 10
DBC = 206
For I and J: For these problems, you use the same concept as before. You have a right angle (XYZ) that has within it two other angles (XYW and WYZ). A right angle has 90 degrees. Combine the two unknown angles and set it equal to the right angle. Solve for x.
XYW + WYZ = XYZ
(1 1/4x - 10) + (3/4x + 20) = 90
1 1/4x - 10 + 3/4x + 20 = 90
2x + 20 = 90
2x = 70
x = 35
Plug x into the angle values and solve.
XYW = 1 1/4x - 10
XYW = 1 1/4(35) - 10
XYW = 43.75 - 10
XYW = 33.75
WYZ = 3/4x + 20
WYZ = 3/4(35) + 20
WYZ = 26.25 + 20
WYZ = 46.25
Answer:
If corresponding vertices on an image and a preimage are connected with line segments, the line segments are divided equally by the line of reflection. That is, the perpendicular distance from the line of reflection to either of the corresponding vertices is the same. Line is a perpendicular bisector of the connecting line segments.
Step-by-step explanation: