In the first million digits of pi,how many threes are there?
2 answers:
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Answer:
Ratio = 5/3
The ratio (larger to smaller) of the perimeters is 5/3
Step-by-step explanation:
Attached is an image of the two triangles.
Since both triangles are similar, the ratio of their perimeter is equal to the ratio of each similar sides.
Ratio = P1/P2 = S1/S2
Similar side for triangle 1 S1 = 15ft
Similar side for triangle 2 S2 = 9ft
Substituting the values;
Ratio = 15ft/9ft = 5/3
Ratio = 5/3
The ratio (larger to smaller) of the perimeters is 5/3
Answer:
The length of the longer base is 80 feet.
Step-by-step explanation:
We are given that county courthouse is shaped like a trapezoid.
Let the height of the court house be h
Since we are given that The shorter base of the trapezoid is 15 feet longer than the height.
So, the shorter base = h+15
We are also given that The longer base is 5 feet longer than 3 times the height
So, longer Base = 3h+5
Now we are also given that the area of the landing is 1500 square feet.
Formula of area of trapezium :
=
Substituting the values .
Taking 4 common from the equation.
⇒ h = 25, -30
Neglect the negative value .
So, height = 25 feet.
Length of Longer base = 3h+5 =3*25+5=80 feet
Hence , the length of the longer base is 80 feet.
Answer:
a. θ = 30°
b. μ = √3 / 15 ≈ 0.115
Step-by-step explanation:
Draw a free body diagram for each scenario (see attached figure). The body has four forces acting on it:
- Weight pulling down
- Normal force perpendicular to the incline
- Applied force parallel up the incline
- Friction force parallel to the incline
Remember that friction opposes the direction of motion. So when the body is sliding up, friction points down the incline. And when the body is sliding down, friction points up the incline.
Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.
For sliding up, sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
P₁ − f − mg sin θ = 0
P₁ − Nμ − mg sin θ = 0
Substituting the expression for normal force:
P₁ − mgμ cos θ − mg sin θ = 0
Now for sliding down, sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
And sum of the forces parallel to the incline:
∑F = ma
P₂ + f − mg sin θ = 0
P₂ + Nμ − mg sin θ = 0
Substituting the expression for normal force:
P₂ + mgμ cos θ − mg sin θ = 0
We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.
So we have two equations and two unknowns (μ and θ):
P₁ − mgμ cos θ − mg sin θ = 0
P₂ + mgμ cos θ − mg sin θ = 0
Let's start by adding the equations together:
P₁ + P₂ − 2 mg sin θ = 0
P₁ + P₂ = 2 mg sin θ
sin θ = (P₁ + P₂) / (2 mg)
Plugging in the values:
sin θ = (6 + 4) / (2 × 10)
sin θ = 1/2
θ = 30°
Now we can plug this into either equation and find μ.
P₁ − mgμ cos θ − mg sin θ = 0
6 − (10 cos 30°) μ − 10 sin 30° = 0
6 − 5√3 μ − 5 = 0
1 = 5√3 μ
μ = √3 / 15
μ ≈ 0.115
