Its 6$ for 1 pair of pants so for 11 pairs of pant's it is 66$
Answer:
1.7 in.
Step-by-step explanation:
Draw an equilateral triangle with a horizontal side on the bottom. All sides are congruent, and all angles are congruent. Each side measures 2 inches. Each angle measures 60 degrees.
Now draw the altitude from the top vertex to the bottom horizontal side. The altitude is perpendicular to the horizontal side and bisects it. Now you have two right triangles inside the original equilateral triangle. The hypotenuse of each right triangle is a side of the original equilateral triangle. It measures 2 in. Since the bottom 2-inch side of the original equilateral triangle was bisected by the altitude, the short leg of each right triangle is half the length of the side of the equilateral triangle. Therefore, the short leg of each right triangle measures 1 inch. We can find the length of the long leg (the altitude of the equilateral triangle) by using the Pythagorean theorem. Let the unknown height be h.




Take the square root of 3. You'll get 1.73205...
Round it off to the nearest tenth to get 1.7.
Answer: 1.7 in.
20 + 18 = 38
38 is the perimeter
The length of AB is 9.
Solution:
Given data:
Radius OC = 8
Tangent AC = 15
The angle between the tangent and radius is always right angle.
∠C = 90°.
Hence OCA is a right triangle.
Using Pythagoras theorem,
<em>In a right triangle square of the hypotenuse is equal to the sum of the squares of the other two sides.</em>




Taking square root on both sides of the equation, we get
OA = 17
OB is the radius of the circle.
⇒ OB = 8
AB = OA – OB
= 17 – 8
= 9
AB = 9
Hence the length of AB is 9.
The solutions or roots to this equation is found by solving for x. We can do this a couple ways either by FOIL or quad formula
3x^2+6x+6=0
3(x^2+2x+2)=0
x^2+2x+2=0 we cant FOIL this out so we use the quad formula
x= [(-b+\-sqrt(b^2-4ac))/2a]
x= -2+\- sqrt(4-4(1)(2))/2(1)
x= -1 +i , -1 - i
So we have complex roots since our quad formula returned a negative number. Whenever the quad formula answer is positive we have two roots/solutions, when it is zero we have one root/solution, and whenever it is negative we have Complex roots/solutions
Hope this helps. Any questions please just ask. Thank you.