Answer: (k+2) (k+3)
Step-by-step explanation:
Factor 5^2 +5k+6 using the AC method.
Answer:
18,223 feet
Step-by-step explanation:
y= -16t^2 + 608t + 12,447.
y = -16(t² - 38t) + 12447
y = -16[t² - 2(t)(19) + 19² - 19²] + 12447
y = -16(t - 19)² - 16(-19²) + 12447
y = -16(t - 19)² + 18223
Vertex: (19, 18223)
Max height is 18,223 at t = 19
Answer:
![x = 38.1](https://tex.z-dn.net/?f=x%20%3D%2038.1)
![y = 23.4](https://tex.z-dn.net/?f=y%20%3D%2023.4)
Step-by-step explanation:
The given triangle is a right triangle. This is indicated by the box around one of the angles, signifying that the angle measure is (9) degrees. By its definition, a right triangle is a triangle with a (90) degree angle, thus the given triangle is a right triangle. In this situation, one can use right-angle trigonometry to find the value of the unknown side. In essence, right triangle trigonometry provides one with a series of ratios to describe the relationship between the sides and angles of the right trinagle. These ratios are the following,
![sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D%5Cfrac%7Bopposite%7D%7Bhypotenuse%7D%5C%5C%5C%5Ccos%28%5Ctheta%29%3D%5Cfrac%7Badjacent%7D%7Bhypotenuse%7D%5C%5C%5C%5Ctan%28%5Ctheta%29%3D%5Cfrac%7Bopposite%7D%7Badjacent%7D)
Keep in mind that the way a side is named is relative to the angle it is being described from. Thus, the (opposite) and (adjacent) sides acquire different names based on the angle relative to them. However, the (hypotenuse) is the hypotenuse no matter the angle, as the hypotenuse is the side opposite the right angle.
Solve for (y) first. One is given the angle measure and the measure of the side opposite the angle measure. One is asked to find the measure of the side adjacent to the angle. Use the ratio of tangent (tan) to achieve this.
![tan(\theta)=\frac{opposite}{adajcent}\\](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7Bopposite%7D%7Badajcent%7D%5C%5C)
Substitute,
![tan(52)=\frac{30}{y}](https://tex.z-dn.net/?f=tan%2852%29%3D%5Cfrac%7B30%7D%7By%7D)
Inverse operations,
![y=\frac{30}{tan(52)}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B30%7D%7Btan%2852%29%7D)
Simplify,
![y = 23.4386](https://tex.z-dn.net/?f=y%20%3D%2023.4386)
Now solve for the hypotenuse (x). One could use the Pythagorean theorem to do this, but since this is trigonometry, one might as well use the trigonometric ratios. One is given an angle, as well as the side opposite the angle, one is asked to find the hypotenuse. To do this, one can use the trigonometric ratio of sine (sin).
![sin(\theta)=\frac{opposite}{hypotenuse}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D%5Cfrac%7Bopposite%7D%7Bhypotenuse%7D)
Substitute,
![sin(52)=\frac{30}{x}](https://tex.z-dn.net/?f=sin%2852%29%3D%5Cfrac%7B30%7D%7Bx%7D)
Inverse operations,
![x=\frac{30}{sin(52)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B30%7D%7Bsin%2852%29%7D)
Simplify,
![x = 38.0705](https://tex.z-dn.net/?f=x%20%3D%2038.0705)
The count of the equilateral triangle is an illustration of areas
There are 150 small equilateral triangles in the regular hexagon
<h3>How to determine the number of
equilateral triangle </h3>
The side length of the hexagon is given as:
L = 5
The area of the hexagon is calculated as:
![A = \frac{3\sqrt 3}{2}L^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B3%5Csqrt%203%7D%7B2%7DL%5E2)
This gives
![A = \frac{3\sqrt 3}{2}* 5^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B3%5Csqrt%203%7D%7B2%7D%2A%205%5E2)
![A = \frac{75\sqrt 3}{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B75%5Csqrt%203%7D%7B2%7D)
The side length of the equilateral triangle is
l = 1
The area of the triangle is calculated as:
![a = \frac{\sqrt 3}{4}l^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5Csqrt%203%7D%7B4%7Dl%5E2)
So, we have:
![a = \frac{\sqrt 3}{4}*1^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5Csqrt%203%7D%7B4%7D%2A1%5E2)
![a = \frac{\sqrt 3}{4}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5Csqrt%203%7D%7B4%7D)
The number of equilateral triangles in the regular hexagon is then calculated as:
![n = \frac Aa](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%20Aa)
This gives
![n = \frac{75\sqrt 3}{2} \div \frac{\sqrt 3}4](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B75%5Csqrt%203%7D%7B2%7D%20%5Cdiv%20%5Cfrac%7B%5Csqrt%203%7D4)
So, we have:
![n = \frac{75}{2} \div \frac{1}4](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B75%7D%7B2%7D%20%5Cdiv%20%5Cfrac%7B1%7D4)
Rewrite as:
![n = \frac{75}{2} *\frac{4}1](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B75%7D%7B2%7D%20%2A%5Cfrac%7B4%7D1)
![n = 150](https://tex.z-dn.net/?f=n%20%3D%20150)
Hence, there are 150 small equilateral triangles in the regular hexagon
Read more about areas at:
brainly.com/question/24487155