The correct answer for that question is 280
1. The problem statement tells you to find "the area of the hexagonal face".
2. If we assume the intent is to find the shaded area of the face only, it differs from the area of a regular hexagon in that there is a hole in the middle.
3. You must find the area of the regular hexagon, and subtract the area of the circular hole in the middle.
4. The formula for the area of a circle in terms of its radius is
... A = πr²
5. The formula for the area of a regular hexagon in terms of the radius of the circumcircle is
... A = (3√3)/2·r²
6. The radius of the circumcircle of the regular hexagon is given. No additional information is needed.
7. You can use the trig functions of the angles of an equilateral triangle to find the apothem, but there is no need for that when you use the formula of 5.
8. All this is unnecessary. The apothem is (8 mm)·(√3)/2 = 4√3 mm ≈ 6.9282 mm, the shorter leg is (8 mm)·(1/2) = 4 mm. The perimeter is 6·8 mm = 48 mm.
9. The area of the hexagon is
... A = 3√3/2·(8 mm)² = 96√3 mm² ≈ 166.277 mm²
10. The area of the circle is
... A = π·(4 mm)² = 16π mm² ≈ 50.265 mm²
11. The area of the hexagonal face is approximately ...
... 166.277 mm² - 50.265 mm² = 116.01 mm²
I think estimation is terrible. If u want to truly solve it, you just need to get a common denominator on each of the fractions. that means the bottom half of both fractions must be the same in addition or subtraction. so here, the common denominator would be 66. What times 6 = 66? 11. So 1*11/6*11 = 11/66. Then what times 11 = 66? 6. So 6*6/6*11.
Your new problem is 2 11/66 + 3 36/66. Now you can just add it up, but the denominators will always stay the same, 66. Your answer would be 5 47/66
The true statement is h(2)=16.
<h3 /><h3>What is domain and range?</h3>
The domain of a function is the set of values that we are allowed to plug into our function. The range of a function is the set of values that the function assumes.
Given:
domain of -3 ≤ x ≤ 11 and a range of 1 ≤ h(x) ≤ 25,
Also, h(8) = 19 and h(-2) = 2,
Now,
2=h(-2)< h(2)<h(8) =19
h(8)=19≠21
h(13)> h(8) =19
h(-3)< h(-2) =2 [1 ≤ h(x) ≤ 25]
Hence, h(2)=16
Learn more about domain and range:
brainly.com/question/23199615
#SPJ1
Answer:
10.17 seconds
Step-by-step explanation:
substitute y=0 into the equation
0=-16x²+153x+98
then use the quadratic formula
a = -16
b= 153
c = 98
Substituting values into the Quadratic equation, you get (Round the values to the nearest hundreth):
x₁= 10.17
x₂= -0.60
