Answer:
Add equations A + B to eliminate y
Then add equations A and C to eliminate y
Step-by-step explanation:
Since all the equations have y with a coefficient of 2, I would eliminate y
Add equations A + B to eliminate y
Then add equations A and C to eliminate y
Assuming that the given triangle is a right triangle where height and base are equal, we can solve for the missing variables. Let us review the given problem, the given is the total area since unit of "a" is inch squared.
The formula for an area of a triangle is At = 1/2(bxh). We substitute the given value and assume that h=b.
Therefore, 48=1/2bh. To eliminate 1/2, we multiply both sides by 2. Then we have 96=bh. When b=h, we have b^2=96.
The missing variables are b=h=9.8.
Answer:
The answer is (g°f)(-1) = 27
Step-by-step explanation:
* (g°f)(1) ⇒ means make the domain of g is the range of f
- At first find the value of f(-1)
- Take the answer and substitute it as the value of x for g
- So the range of f is the domain of g
∵ f(x) = 4x + 7
∵ The domain of f = -1 ⇒ x is the domain
∴ f(-1) = 4(-1) + 7 = -4 + 7 = 3 ⇒ f(-1) is the range
∵ g(x) = x³
∵ x = f(-1) = 3 ⇒ the domain of g
∴ g(3) = 3³ = 27 ⇒ the range of g
* (g°f)(-1) = 27
When we divide the figure in four parts, we obtain four squares with its sides of 5 centimeters of lenght, with quarter circles (two) in each one of them. The limits of the shadded area are arcs of cirncunference
To calculate the area of the shadded part, we must choose one square of 5cmx5cm and divide it in 3 sectors:
a: the area below the shaded area.
b: the area above the shaded area.
c: the shaded area.
The area of the square (As) is:
As=L²
As=(5 cm)²
As=25 cm²
The area of a circunference is: A=πR² (R:radio), but we want the area of the quarter circle, so we must use A=1/4(πR²), to calculate the area of the sectors a+c:
A(a+c)=1/4(π(5)²)
A(a+c)=25π/4
The area of the sector "b" is:
Ab=As-A(a+c)
Ab=25-25π/4
Ab=25(1-π/4)
The area of the sector a+b, is:
A(a+b)=2Ab
A(a+b)=2x25(1-π/4)
A(a+b)=50(1-π/4)
Then, the shadded area (Sector c) is:
Ac=As-A(a+b)
Ac=25-50(1-π/4)
Ac=25-(50-50π/4)
Ac=25-50+50π/4
Ac=(50π/4)-25
Ac=(25π/2)-25
Ac=25(π/2-1)
The area of each shaded part is: 25(π/2-1)
To calculate the perimeter of a shaded part, we must remember that the perimeter of a circunference is: P=2πR. If we want the perimeter of a quarter circle we must use: P= 2πR/4. But there is two quarter circles in the square of 5cmx5cm, so the perimeter of the shaded area is:
P=2(2πR/4)
P=4πR/4
P=πR
P=5π
The perimeter of each shaded part is: 5π
