1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP
If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8
2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2
Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4
FC= 2/(1+2) * AC
FC= 2/3 * 12= 8
3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2
DC=3x+12= 3(-2) +12= 6
4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3
C
BC lines slope would be (-1)-(-1)/4 = 0/4.
The line would be
0=x+a
a=-1
0=x-1
x=1
y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
x=1
y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)
5.
a. Circumcenter: the intersection of perpendicular bisector lines<span>
b. Incenter: the intersection of bisector lines
c. Centroid: </span>the intersection of median lines<span>
d. Orthocenter: </span>the intersection of altitude lines
16. 5x^3 y^-5 • 4xy^3
20x^4y^-2
20x^4 • 1/y^2
=20x^4/y^2
17. -2b^3c • 4b^2c^2
= -8b^5c^3
18. a^3n^7 / an^4 (a^3 minus a = a^2 same as n^7 minus n^4 = n^3)
=a^2n^3
19. -yz^5 / y^2z^3
= -z^2/y
20. -7x^5y^5z^4 / 21x^7y^5z^2 (divide -7 to 21 and minus xyz)
= -z / 3x^2
21. 9a^7b^5x^5 / 18a^5b^9c^3
=a^2c^2 / 2b^4
22. (n^5)^4
n ^5 x 4
=n^20
23. (z^3)^6
z ^3 x 6
=z^18
Answer:
4
Step-by-step explanation:
B)
2x6= 12
2x7= 14
2x8= 16
2x8= 18
There is no picture what do you mean
