Answer:
0.44
Step-by-step explanation:
-3x^2 + 2y^2 + 5xy - 2y +5x^2 - 3y^2
Combine like terms
-3x^2 + 5x^2 = 2x^2 2y^2 - 3y^2 = -1y^2
2x^2 - 1y^2 + 5xy - 2y
Now plug in the solutions Note: it is easier if you have all decimals or all fractions (-1/10=-.1
2(0.5)^2 - 1(-0.1)^2 + 5(0.5)(-0.1) - 2(-0.1)
Simplify:
0.5 - 0.01 - 0.25 + 0.2
0.5 + 0.2 - 0.01 - 0.25
0.7 - 0.26
0.44
I think it’s-10. Hope it’s right.
<span>Dawn was at 6 am.
Variables
a = distance from a to passing point
b = distance from b to passing point
c = speed of hiker 1
d = speed of hiker 2
x = number of hours prior to noon when dawn is
The first hiker travels for x hours to cover distance a, and the 2nd hiker then takes 9 hours to cover that same distance. This can be expressed as
a = cx = 9d
cx = 9d
x = 9d/c
The second hiker travels for x hours to cover distance b, and the 1st hiker then takes 4 hours to cover than same distance. Expressed as
b = dx = 4c
dx = 4c
x = 4c/d
We now have two expressions for x, set them equal to each other.
9d/c = 4c/d
Multiply both sides by d
9d^2/c = 4c
Divide both sides by c
9d^2/c^2 = 4
Interesting... Both sides are exact squares. Take the square root of both sides
3d/c = 2
d/c = 2/3
We now know the ratio of the speeds of the two hikers. Let's see what X is now.
x = 9d/c = 9*2/3 = 18/3 = 6
x = 4c/d = 4*3/2 = 12/2 = 6
Both expressions for x, claim x to be 6 hours. And 6 hours prior to noon is 6am.
We don't know the actual speeds of the two hikers, nor how far they actually walked. But we do know their relative speeds. And that's enough to figure out when dawn was.</span>
Answer:
no, it's a mcdonalds pepsi
Step-by-step explanation:
