Answer:
The answer to your question is AC = 9.9
Step-by-step explanation:
To solve this problem use trigonometric functions. The trigonometric function that relates the hypotenuse and the adjacent side is cosine.
cos α = 
Solve for hypotenuse

Substitution
hypotenuse = 
Simplification and result
hypotenuse = 9.85
Answer:
for the die it's a 4 to 6 chance and the coin is a 50% chance
Answer:
yeah why
<u>Step-by-step explanation:</u>
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The percentage of the semicircle shaded section is approximately 23,606 %.
The percentage of the area of the semicircle is equal to the ratio of the semicircle area minus the half-cross area to the semicircle area. In other words, we have the following expression:

(1)
Where:
- Area of the half cross, in square centimeters.
- Area of the semicircle, in square centimeters.
- Percentage of the shaded section of the semicircle.
And the percentage of the shaded section is:
![r = \left[1-\frac{4 \cdot (2\,cm)^{2}+4\cdot \left(\frac{1}{2} \right)\cdot (2\,cm)^{2}}{0.5\cdot \pi\cdot (16\,cm^{2}+4\,cm^{2})} \right]\times 100](https://tex.z-dn.net/?f=r%20%3D%20%5Cleft%5B1-%5Cfrac%7B4%20%5Ccdot%20%282%5C%2Ccm%29%5E%7B2%7D%2B4%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5Ccdot%20%282%5C%2Ccm%29%5E%7B2%7D%7D%7B0.5%5Ccdot%20%5Cpi%5Ccdot%20%2816%5C%2Ccm%5E%7B2%7D%2B4%5C%2Ccm%5E%7B2%7D%29%7D%20%5Cright%5D%5Ctimes%20100)

The percentage of the semicircle shaded section is approximately 23,606 %.
We kindly invite to check this question on percentages: brainly.com/question/15469506
Answer:
units.
Step-by-step explanation:
Let x be the width of rectangle.
We have been given that the length of garden is 2 units more than 1.5 times it’s width. So length of the rectangle will be:
.
To find the length of total fencing we need to figure out perimeter of rectangle with width x and length
.
Since we know that perimeter of a rectangle is two times the sum of its length and width.

Upon substituting length and width of garden in above formula we will get,


Upon using distributive property we will get,


Therefore, the length of required fencing will be
units.