D- 4 is your correct answer :)
Hope this helps
Answer:
The system has no solution
Step-by-step explanation:
we have
y=5x+10 -----> equation C
The slope of the equation C is m=5 and the y-intercept is b=10
y=5x+2 -----> equation D
The slope of the equation C is m=5 and the y-intercept is b=2
Remember that
If two lines are parallel, then their slopes are the same
Equation C and equation D are parallel lines with different y-intercept
therefore
The system has no solution (the lines do not intersect)
So it can be easier to solve your problem
Answer
a) PQ = 50
b) PC = 18
PY = 27
c) ZC = 6
ZQ = 18
Explanation
Centroid is a point inside a triangle where all lines drawn from each vertex to the midpoints of opposite side intersect.
Part a
X is the midpoint of PQ. SoPX = XQ.
If PX = 25,
then PQ = 2PX
PQ = 2×25
PQ = 50
Part b
The centroid is alway two thirds away from the vertex.
Also PC = twice CY
PC = 2×9
PC = 18
So, if CY = 9, then PY = 3 × 9
PY = 27
Part c
(i) The centroid is alway two thirds away from the vertex.
If ZC = 12, then CZ =(1/2)QZ
CZ = (1/2) × 12
CZ = 6
(ii) The centroid is alway two thirds away from the vertex.
So, if CZ = 6, then QZ =3 CZ
QZ = 3×6
QZ = 18
After 1st year: 250$:100%=x$:116%, 250$*116%=x$*100%, x=(250*116)/100=290$. After 1st year I will have 290$
After 2nd year: 290$:100%=x$:116%, x=(290*116)/100=336.4$. After 2nd year I will have 336.4$
After 3rd year I will have (336.4*116)/100=390.224$
After 4th yr: (390.224*116)/100=452.65984$
After 5th yr: (452.65984*116)/100=525.085$
After- 6th yr: 609.1$, 7th yr: 706.556$, 8th yr: 819.605$, 9th yr: 950.742$
10th yr: 1102.86$, 11th yr: 1279.32$, 12th yr: 1484.01$, 13th yr: 1721.45$,
14th yr: 1996.88$, 15th: 2316.38$, 16th yr: 2687$, 17th yr: 3116.92$
After 18 years I will have 3615.63$.