Answer:
5.0 ft - 5.6 ft
Step-by-step explanation:
Given that the structure is to be made using two 12 foot boards, then we expect the total perimeter to be equal to (2*12)= 24 ft.
Using the angle of elevation, 40° and the width of 8 ft then you can apply the formula for tangent of a triangle where ;
Tan α = opposite side length/adjacent length
Tan 40°= h/8
h= 8 tan 40° = 6.71 ft
Applying the cosine of an angle formula to find the length of the sliding side
Cosine β = adjacent length /hypotenuse
Cosine 40°= 8/ sliding side length
sliding side length = 8/cosine 40° =10.44 ft
Checking the perimeter = 10.44 +8+6.71= 25.15 ft
This is more than the total lengths of the boards, so you need to adjust the height as;
24 - 18.44 = 5.56 ft ,thus the height should be less or equal to 5.56 ft
h≤ 5.6 ft
Answer:
Step-by-step explanation:
He could use either. It would probably be better to use one that already has a variable without a coefficient. which one has that? Let me know if that doesn't make sense.
Answer:
For red you have a 1/5% chance for pink you have a 2/4% chance also know as 50% chance
Step-by-step explanation:
Integrals calculate the area underneath the curve. So, look at the graph, note where the function is on the graph and how many units on the X-axis the function is wide and how many units on the Y-axis the function is tall. The x-axis is usually where you would set the boundaries of the function. The integrand (the thing that we integrate) is the the function typically, and it’s in respect to something else, dx
, dy or dt to name a few. The thing we integrating with respect to, is dependent upon the graph.
Here’s an example: (Picture at the bottom)
If I take my bounds from 0 to 2 on the X-axis, that would give me an area underneath the curve that is exactly like a triangle. You can solve it with half base times height if you wanted.
If I were to take my bounds to be 2 to 6 on the X-axis then that’s an area in the shape of a trapezoid. Remember that you’re integrating the function so that means that your integrand will be the function.
So, for the first one it would be the integral sign, 2 on the top, 0 on the bottom, the integrand would be f(x), and it’s in respect to dt.
For the second one it would be exactly the same except your bounds would just be from 2 to 6
(2 on the bottom 6 on top).
I hope that helps.