Answer:
Step-by-step explanation:
The terms of the sequence are:
We can rewrite the terms in factored form to get;
We can see that the subsequent terms are obtained by multiplying the previous term by 
. This is called the common ratio.
Therefore the first term of this geometric sequence is 
and the common ratio is 
.
The nth term of a geometric sequence is given by: 
.
Let us substitute the first term, the common ratio, and 
to obtain:
The answer is C) 58.5 sq. units
Just count the whole unit squares within the shape and estimate the partial ones and add them all together to get the estimated area of the irregular shape. There were 57 full squares and 3 halves for a total of 58.5 units.
The trigonometric function gives the ratio of different sides of a right-angle triangle. The given problems can be solved as given below.
<h3>What are Trigonometric functions?</h3>
The trigonometric function gives the ratio of different sides of a right-angle triangle.
where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.
1st.) x = 5 /Sin(30°)
x = 10
!) sin(45°) = 4/x
x = 4/sin(45°)
x = 4√2
I) Cos(45°) = √3 / x
x = √3 / Cos(45°)
x = √6
E) Tan(60°) = 3√3 / x
x = 3√3 / 3
W) For isosceles right-triangle, the angle made by the legs and the hypotenuse is always 45°.
x = 45°
N) x² + x² = (7√2)²
x = 7
V) Tan(60°) = 7 / x
x = 7√3/3
K) x² + x² = (9)²
x = 9/√2
Y) Sin(60°) = 7√3/x
x = 14
M) Sin(30°) = x/11
x = 11/2
T) Sin(45°) = x/√10
x = √5
A) x + 2x + 90° = 180°
x = 30°
O) Sin(45°) = √2 / x
x = 2
R) Tan(30°) = x / 4
x = 4/√3 = 4√3 / 3
S) Sin(60°) = x / (10/3)
x = 5√3 / 3
Learn more about Trigonometric functions:
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We would draw 3/4 next to 1/8, and see that 3/4 is 6 times the size of 1/8 so we would see that 3/4 = 6/8, so 6/8+1/8=7/8
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)
