The value of x from the diagram is 29 degrees.
<h3>SOH CAH TOA identity</h3>
From the given figure, we have the following parameters:
- Opposite = 12
- Hypotenuse = 25
<h3>According to the theorem:</h3>
Sin x = opposite/hypotenuse
Sin x = 12/25
Sin x = 0.48
x = arcsin(0.48)
x = 28.685
x = 29 degrees
Hence the value of x from the diagram is 29 degrees.
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Answer:
For maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Step-by-step explanation:
Let the length of the rectangular exercise run = l ft
and width of the run = w ft
Sinoman has to cover a rectangular exercise run from three sides with the fencing material,
So length of the material = (l + 2w) ft
l + 2w = 100
l = 100 - 2w --------(1)
Area of the rectangular area covered = Length × width
A = lw
A = w(100 - 2w) [(l = 100 - 2w)from equation (1)
For maximum area we find the derivative of area and equate it to zero.
![\frac{dA}{dw}=\frac{d}{dw}[w(100-2w)]](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdw%7D%3D%5Cfrac%7Bd%7D%7Bdw%7D%5Bw%28100-2w%29%5D)

A' = 100 - 4w
For A' = 0
100 - 4w = 0
4w = 100
w = 25 ft
From equation (1)
l = 100 - 2w
l = 100 - 2×(25)
l = 50 ft
Therefore, for maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Answer:
The answer is <u>D. quadrilaterals</u>
Step-by-step explanation:
- <u><em>All </em></u><u><em>trapezoids </em></u><u><em>are squares. If a shape is a square, then it is also a rectangle and a rhombus. A</em></u><u><em> trapezoid </em></u><u><em>is a </em></u><u><em>quadrilateral </em></u><u><em>with exactly one set of parallel sides</em></u>
Answer:
Circumference of the circle is 
Step-by-step explanation:
we have to find the circumference of the given circle.
Given:
Diameter=42 m

Circumference of a circle= 

So, the circumference of the circle is 