The convex heptagon has 14 distinct diagonals can be drawn
Step-by-step explanation:
A polygon is said to be a heptagon if it has 7 vertices, 7 sides and 7 angles. A heptagon is called a convex heptagon if the lines connecting any two non-adjacent vertices lie completely inside the heptagon
The formula of number of diagonals in any polygon is 
, where
- d is the number of the diagonals of the polygon
- n is the number of sides of the polygon
∵ The heptagon has 7 sides
∴ n = 7
∵ The number of diagonals = 
- Substitute n by 7 in the rule above
∴ The number of diagonals = 
∴ The number of diagonals = 
∴ The number of diagonals = 
∴ The number of diagonals = 14
The convex heptagon has 14 distinct diagonals can be drawn
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Answer:
The first one
Step-by-step explanation:
2y + 4x = -5
2y = -5 - 4x
y = - (5/2) - 2x
Area of a rectangle is length x width.
Let the width = x
The length would be x +7. ( 7 inches longer than the width)
Area = 20
Set up the formula:
20 = x * x+7
Simplify the right side:
20 = x^2 + 7x
Subtract 29 from both sides:
X^2 + 7x -20 = 0
Solve using the quadratic equation
X = -b + sqrt(b^2 -4ac) / 2a
X = -7 + sqrt(7^2-4(1)(-20) / 2(1) (exact answer)
X = 2.178908 ( as a decimal)
The width is 2.178907 inches (round as needed)
The length would be 9.178907 inches ( round as needed.)
Depending on how you round, when you multiply them together you get approximately 20 square inches.
Compute the definite integral:
integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx
Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
= integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx
Integrate the sum term by term and factor out constants:
= 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Evaluate the antiderivative at the limits and subtract.
(5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand 1/(x^2 + 3 x + 2), complete the square:
= (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx
For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds
Factor -1/4 from the denominator:
= (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds
Factor out constants:
= (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds
Factor -1 from the denominator:
= (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds
For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
= (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp
Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
= (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5
Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
= (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)
Which is equal to:
Answer: = log(18)
If you divide 12 by 384 and put it in a calculator you get 32.this means 12 goes 32 times in 384
