Answer:
Base of the mirror is 12 feet and height of the mirror is 18 feet.
Step-by-step explanation:
Given:
Area of the triangular mirror = 108 sq ft.
Let base (b) of the mirror be x.
Now given:
height is 6 feet less than twice the base of the mirror
So we can say that;
Height (h) = 
We need to find the base and height of the mirror.
Solution:
Now we know that mirror is in triangular shape so we will apply area of triangle formula to find base and height.
Now Area of triangle is half times base times height.
framing in equation form we get;
Now we will find the roots of x by factorizing the equation we get;
Now we will solve for 2 values of x we get;
Now we get 2 values of 'x' one positive and one negative.
Since x is base of the triangle which cannot be negative.
so we will discard the negative value.
Hence;
base of triangle = 12 ft
Height of the triangle = 
Hence Base of the mirror is 12 feet and height of the mirror is 18 feet.
<span>The correct
answer between all the choices given is the last choice or letter D, which is 1570 m3. I am
hoping that this answer has satisfied your query and it will be able to help
you in your endeavor, and if you would like, feel free to ask another question.</span>
Answer:
for bus A if there is OVER 40 studentsbus b will cost less
a^2 evaluates to
a^2+a evaluates to
Multiply the exponent of a by 2giving
The answer is
a^2 evaluates to
+ =
The answer is
a^2+a+a^2 evaluates to
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
