Answer:
Step-by-step explanation:
Cos(A + B) = Cos(A)*Cos(B) - Sin(A)*Sin(B)
Find The Hypotenuse of A
Tan(A) = Opposite / Adjacent
Tan(A) = 15/8
Opposite = 15
Adjacent = 8
Hypotenuse = c
c^2 = opposite^2 + Adjacent^2
c^2 = 15^2 + 8^2
c^2 = 225 + 64
c^2 = 289
sqrt(c^2) = sqrt(289)
c = 17
Find the opposite of B
Adjacent = 12
Hypotenuse = 37
Opposite = b
adjacent^2 + opposite^2 = hypotenuse^2
12^2 + b^2 = 37^2
144 + b^2 = 1369
b^2 = 1369 - 144
b^2 = 1335
b = 35
Definitions
Sin(A) = Opposite / hypotenuse = 15/17
Sin(B) = opposite / hypotenuse = 35/37
Cos(A) = 8/17
Cos(B) = 12/37
Answer
Cos(A + B) = Cos(A)*Cos(B) - Sin(A)*Sin(B)
Cos(A + B) = 8/17 * 12/37 - 15/17*35/37
Cos(A + B) = 96/629 - 525/629
Cos(A + B) = -429/629
We have been given that the diameter of a smaller circle is 7 km and the width of the circular path is 0.5 km. We are asked to find the area of the path.
The area of the path will be equal to area of larger circle minus area of smaller circle.
, where r represents radius of circle.
We know that radius is half the diameter, so radius of smaller circle would be 
km.
The radius of the larger circle will be radius of smaller circle plus width of the path that is 
km.
Therefore, the area of the circular path is approximately 11.8 square km.
Answer: $81.60
Step-by-step explanation:
When we have an original price OP, and a markup of X% (this is how much increases the price with respect to the original price)
The retail price will be:
RP = OP + (X%/100%)*OP
RP = OP*(1 + X%/100%)
In this case, we have:
Original price = OP = $68
markup = X% = 20%
Replacing these in the above equation, we get:
RP= $68*(1 + 20%/100%) = $68*(1 + 0.20)
RP = $68*(1.20) = $81.60
Then the retail price of this particular item is $81.60
The type of transformations that maps triangle ABC onto Triangle prime ABC is a Reflection across the line y=x ; translation 10 units to the right and 4 units up.
<h3>What is the reflection about?</h3>
Note that: Triangle ABC has vertices at points which are:
A(-6,2), B(-2,6) and C(-4,2).
Therefore, the reflection across the line y=x has the rule of"
(x, y) ---(y, x).
Hence:
A(-6,2) -- A''(2,-6);
B(-2,6)--- B''(6,-2);
C(-4,2)----C''(2,-4).
2. The translation 10 units to the right and 4 units up is:
(x, y)----(x+10,y+4).
Hence
A''(2,-6)----A'(12,-2);
B''(6,-2)----B'(16,2);
C''(2,-4)----C'(12,0).
Therefore, Points A'B'C' are said to be exactly of the vertices of that of triangle A'B'C'.
Hence, The type of transformations that maps triangle ABC onto Triangle prime ABC is a Reflection across the line y=x ; translation 10 units to the right and 4 units up.
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