If m varies directly as p, and m=5 when p=7, what is the constant of variation?
2 answers:
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See the attached picture to better understand the problem
we know that
<span>the area of the shaded portion in the circle=area 1+area 2
step 1
find area 1
area 1=area of semi circle
area 1=pi*r</span>²/2
diameter=12 units
radius r=12/2----> 6 units
r=6 units
area 1=pi*6²/2----> 56.52 units²
step 2
find the area 2
area 2=area sector CADC-area triangle ACD
in the right triangle ABC
BC=3
AC=r-----> AC=6
AB=?
applying Pythagoras Theorem
AC²=AB²+BC²------> AB²=6²-3²-----> AB²=27----> AB=3√3 units
area triangle ACD=b*h/2
b=2*3√3---> 6√3 units
h=3 units
area triangle ACD=6√3*3/2----> 9√3 units²-----> 15.59 units²
find the angle ACB
tan ∠ACB=AB/BC-----> 3√3/3---> √3
∠ACB=arc tan (√3)-----> 60°
the central angle ACD=60*2-----> 120°
area of sector CADC=(120/360)*pi*r²----> (120/360)*pi*6²---> 37.68 units²
area 2=area sector CADC-area triangle ACD
area 2=37.68-15.59----> 22.09 units²
step 3
the area of the shaded portion in the circle=area 1+area 2
the area of the shaded portion in the circle= 56.52 +22.09----> 78.61 units²
the answer is
the area of the shaded portion is 78.61 units²
we know that
<span>the area of the shaded portion in the circle=area 1+area 2
step 1
find area 1
area 1=area of semi circle
area 1=pi*r</span>²/2
diameter=12 units
radius r=12/2----> 6 units
r=6 units
area 1=pi*6²/2----> 56.52 units²
step 2
find the area 2
area 2=area sector CADC-area triangle ACD
in the right triangle ABC
BC=3
AC=r-----> AC=6
AB=?
applying Pythagoras Theorem
AC²=AB²+BC²------> AB²=6²-3²-----> AB²=27----> AB=3√3 units
area triangle ACD=b*h/2
b=2*3√3---> 6√3 units
h=3 units
area triangle ACD=6√3*3/2----> 9√3 units²-----> 15.59 units²
find the angle ACB
tan ∠ACB=AB/BC-----> 3√3/3---> √3
∠ACB=arc tan (√3)-----> 60°
the central angle ACD=60*2-----> 120°
area of sector CADC=(120/360)*pi*r²----> (120/360)*pi*6²---> 37.68 units²
area 2=area sector CADC-area triangle ACD
area 2=37.68-15.59----> 22.09 units²
step 3
the area of the shaded portion in the circle=area 1+area 2
the area of the shaded portion in the circle= 56.52 +22.09----> 78.61 units²
the answer is
the area of the shaded portion is 78.61 units²
Answer:
The distance is 5.20m
Step-by-step explanation:
Here, we want to get the height of the ladder
To get this, we need to understand that we have a right triangle, with the base of the triangle as 3 m, the angle at the downward part is 60, and we want to get the height of the triangle
The angle here faces the height we want to calculate
This height is known as the opposite
What we have is the adjacent
Now, the trigonometric identity that relates the opposite and the adjacent is the tan
The tangent of an angle is the ratio of the opposite to the adjacent
Mathematically;
tan 60 = h/3
h = 3 tan 60
h = 5.20 m
The ladder is at a height of 5.20 m above the ground
