Answer:
41.27cm²
Step-by-step explanation:
4²π=50.2654824574
3×3=9
50.2654824574-9=41.2654824574
41.27cm²
<h3>
Answer: Lower right corner (ie southeast corner)</h3>
In this graph, it is impossible to draw a single straight vertical line through more than one point on the yellow line. Therefore, we conclude that this graph passes the vertical line test, which indicates we have a function.
In contrast, the upper left corner fails the vertical line test. Note the left-most pair of points are vertically stacked together. A single vertical line goes through these two points. So this is one possible way to show the graph does not pass the vertical line test, and thereby making this not a function. The upper right corner and lower left corner has the same idea as the upper left corner, so they aren't functions either.
Answer:
1) x = 6.6.
2) x = 6.1.
Step-by-step explanation:
1) In this triangle, the hypotenuse is given, which is 8 units, and the angle is given, which is 35 degrees. The base is unknown. To find the base, following ratio will be used:
cos θ = Base/Hypotenuse.
cos 35 = x/8.
x = 8*cos 35.
x = 6.6 units (to the nearest tenth)!!!
2) In this triangle, the base is given by 12, and the angle is given by 27 degrees. The perpendicular is unknown. To find the perpendicular in this case, tangent formula will be used:
tan θ = Perpendicular/Base
tan 27 = x/12.
x = 12*tan 27.
x = 6.1 units (to the nearest tenth)!!!
-1 is the first line after 0
8 is 3rd line one the right side after 5
-11 is just first line after 10..
I hope U understand
Please express the eqn of the parab by <span>y = 2x^2. " ^ " indicates exponentiation.
Find the equation of the tangent line to </span>y = 2x^2 at (1,2):
dy/dx = 4x; now let x = 1. Thus, the slope of the TL at (1,2) is 4(1) = 4, and the eqn of the TL is y-2 = 4(x-1), or y-2 = 4x -4, or y = 4x -2.
To find the area bounded by the parabola, the line y = 4x-2 and the x-axis, we need to use horizontal strips and integrate with respect to y instead of to x.
The smallest y value will be 0 (this is the x-axis), and the largest will be 2 (this comes from the given point, (1,2) ).
Solve y = 2x^2 for x: x^2 = y/2, so that x = sqrt(y/2) for y = 0 to y = 2.
y+2
Solve y = 4x - 2 for x: 4x = y + 2, so that x = -------
4
Thus, the length of each horiz. strip of width dy is given by
(y+2)
------- - sqrt(y/2), since the first term represents the larger x-value and the
4 second term represents the smaller x-value.
Then the area is the integral from y=0 to y=2 of
(y+2)
[ -------- - sqrt(y/2) ] dy
4
Can you do this integration? If not, ask specific questions so that I could help you further.