9514 1404 393
Answer:
D. (4, -6)
Step-by-step explanation:
Since the heights of both triangles are the same (CD), their areas are proportional to their bases. That is ...
AC : CB = 3 : 4
In terms of coordinates, ...
4(C -A) = 3(B -C) . . . . multiply the ratios by 4(CB)
4C -4A = 3B -3C . . . .simplify
7C = 4A +3B . . . . . . . isolate C
C = (4A +3B)/7 . . . . . divide by 7
C = (4(1, -9) +3(8, -2))/7 = (28, -42)/7 . . . . substitute given coordinates
C = (4, -6)
I'll answer 3 of these questions.
For future notice, make sure to limit yourself to asking at most, 3 individual questions per question. :)
1) r/10+4=5
Subtract both sides by 4.
(r/10+4)-4=(5)-4
r/10=1
Multiply both sides by 10.
(r/10)*10=(1)*10
r=10
2) n/2+5=3
Subtract 5 from both sides.
(n/2+5)-5=(3)-5
n/2=-2
Multiply both sides by 2.
(n/2)*2=(-2)*2
n=-4
3) 3p-2=-29
Add 2 to both sides.
(3p-2)+2=(-29)+2
3p=-27
Divide both sides by 3.
(3p)/3=(-27)/3
p=-9
Hope this helps.
-Benjamin
Answer:
At a combined speed of 6 in/min, it takes us 24 mins to clean the wall
Step-by-step explanation:
Since the question did not provide the speed with which each student cleans, we can make assumptions. This is so that we can solve the question before us
Assuming student 1 cleans at a speed of 2 inches per minute, student 2 cleans at a speed of 2½ inches per minute & student 3 cleans at a speed of 1½ inches per minute.
Let's list the parameters we have:
Height of wall (h) = 12 ft, Speed (student 1) = 2 in/min, Speed (student 2) = 2½ in/min, Speed (student 3) = 1½ in/min
Speed of cleaning wall = Height of wall ÷ Time to clean wall
Time to clean wall (t) = Height of wall ÷ Speed of cleaning wall
since students 1, 2 and 3 are working together, we will add their speed together; v = (2 + 2½ + 1½) = 6 in/min
1 ft = 12 in
Time (t) = h ÷ v = (12 * 12) ÷ 6 = 144 ÷ 6
Time (t) = 24 mins
Answer:
<u>x</u>
Step-by-step explanation:
Let's simplify the LHS so it can be equated to the RHS.
Hence, the missing term in the numerator of the answer is <u>x</u>
The answer is C I’m pretty sure
