Let the speed for the first 12 mi be x mi/h, the speed for 18 mi was (x+4) mi/h
thus given that
time=distance/speed
the average time will be:
3=(12+18)/(x+x+4)
3=30/(2x+4)
solving for x we get
3(2x+4)=30
6x+6=30
6x=24
x=4 mi/hr
Answer: 12 mi/hr
Answer:
In this equation, we can start by understanding that "x" has a value of 8, as given in the ordered pair. When multiplied by 5, this leads to "40 - 2y = 30". Next, we can subtract 40 from both sides of the equation. This leads us to a value of "-2y = -10". The next step would be to divide both sides by -2 as a way of isolating "y", which leads us to a final value of "y = 5". The final ordered pair would be (8,5).
Answer:
Step-by-step explanation:
B(2,10); D(6,2)
Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)
Rhombus all sides are equal.
AB = BC = CD =AD
distance = √(x2-x1)² + (y2- y1)²
As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x
A(X,0)
AB = AD
√(2-x)² + (10-0)² = √(6-x)² + (2-0)²
√(2-x)² + (10)² = √(6-x)² + (2)²
√x² -4x +4 + 100 = √x²-12x+36 + 4
√x² -4x + 104 = √x²-12x+40
square both sides,
x² -4x + 104 = x²-12x+40
x² -4x - x²+ 12x = 40 - 104
8x = -64
x = -64/8
x = -8
A(-8,0)
Let C(a,b)
M is AC midpoint
(-8+a/2, 0 + b/2) = M(4,6)
(-8+a/2, b/2) = M(4,6)
Comparing;
-8+a/2 = 4 ; b/2 = 6
-8+a = 4*2 ; b = 6*2
-8+a = 8 ; b = 12
a = 8 +8
a = 16
Hence, C(16,12)
Answer:
- 4.2 degrees
Step-by-step explanation:
If the drop was from temp 0 at 8 pm to temp -16.8 at midnight (4 hours later)
then,the drop is given by the rate: -16.8/4 = - 4.2 degrees per hour
Therefore, at 9 pm the temperature was:
0 degrees - 4.2 degrees = - 4.2 degrees
Answer:
i hope you understand this
