We know that V = a^3 and that it's 1000 cubic meters (or unit cubes as long as one unit cube is 1x1x1m). We can write the calculation as attached below. The result (dimensions of one cube house) is
10x10x10 meters.
2/3 + 1/2 is 7/6
Hope this helps!
Answer:
( 6, pi/6)
Step-by-step explanation:
( 3 sqrt(3), 3)
To get r we use x^2 + y ^2 = r^2
( 3 sqrt(3) )^2 + 3^2 = r^2
9 *3 +9 = r^2
27+9 = r^2
36 = r^2
Taking the square root of each side
sqrt(36) = sqrt(r^2)
6 =r
Now we need to find theta
tan theta = y/x
tan theta = 3 / 3 sqrt(3)
tan theta = 1/ sqrt(3)
Taking the inverse tan of each side
tan ^-1 ( tan theta) = tan ^ -1 ( 1/ sqrt(3))
theta = pi /6
Hello!
First of all, we can subtract the stretching time. This gives us 20. If we divide by the four laps we get 5 minutes per lap.
Now, one lap is 400 meters (most tracks are), which is equal to 15,748.03 inches. This means it takes her five minutes to walk 15,748.03 inches. This is also equal to 300 seconds, so it takes her 300 minutes per 15,748.03 inches.
But if we round our big inches number to the nearest ten thousandth, we get 16,000, so in a simpler form her pace is 300/16,000. But we need to find in per second. Therefore we will divide by 300 to find how many inches she walks per second. This means she walks about 53.33 inches per second.
I hope this helps!
We solve the inequality by subtracting 56.50 from both sides of the equation,
10.45b + 56.50 - 56.50 < 292.67 - 56.50
10.45b < 236.17
Then, divide both sides of the inequality by 10.45
b < 22.6
The solution suggests that the number of boxes than can be loaded on a truck without exceeding the weight limit of the truck should always be lesser than 22.6. Since we are talking about number of boxes, the maximum number of boxes that can be loaded should only be 22.
