(20:4)x3=15 (Ellas grade)
(20:5)x4=16 (Minhs grade)
Answer:
55 deg 30 Minutes = 55.5 degrees
sin = opp/hyp
sin(55.5) = altitude/10
Altitude = 10*sin(55.5)
Altitude = 10 * sin(55.5) = 8.241261886
Rounded to nearest 10th
Altitude = 8.2 cm
b. for base use Low of Cosines
Let C = base
Side A and B = length 10 Cm
c^2 = 2*a^2 -2a^2*cos(<C)
<C = 180 - 2*55.5 = 180 -111 = 69 degrees
c = sqrt ( 2a^2 - 2a^2*cos(<C)
c = sqrt ( 2*100- 200*cos(69))
c = sqrt (200 - 200*cos(69) ) = 11.32812474
Answer base
c = 11.3 cms
c.
Area = (1/2)*b * h
Area = (1/2)*(answer a * answer b)
Area = (1/2)* 11.32812474* 8.241261886 = 46.67902132
Area = 46.7 (rounded)
Area = ( 1/2) * a*b*sin(<C) = (1/2)*10*10*sin(69) =
Area = 50* sin(69) = 46.67902132
Area = 46.7 (rounded)
Step-by-step explanation:
Answer:
The correct answer is: Option 3.
Step-by-step explanation:
To begin the problem asks to find 
· 
which can be defined as ](https://tex.z-dn.net/?f=%5Bf%2Ag%5D%28x%29)
. Now we are given that:
So now we need to 'multiply' our two algebraic expressions as follow:
Eqn.(1)
The domain of Eqn. (1) is all real numbers of 
.
Which according to the given options , Option 3 is correct.
Step-by-step explanation:
It is given that,
The length of section 1 is 0.75 m and the length of section 2 is 15/100 m. The total length of the section is equal to the sum of section 1 and section 2. So, we will add them as follows :
So, the length of the section is 0.9 m or 9/10 m
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