The sum of the given series can be found by simplification of the number
of terms in the series.
- A is approximately <u>2020.022</u>
Reasons:
The given sequence is presented as follows;
A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021
Therefore;
The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;
Therefore, for the last term we have;
2 × 2043231 = n² + 3·n + 2
Which gives;
n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0
Which gives, the number of terms, n = 2020
Which gives;
Learn more about the sum of a series here:
brainly.com/question/190295
Answer:
1.A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.
2.
Solution given:
We have
Sin²
-Cos ²=1
Sin²=1+Cos²
Sin=
3.
4Tan²+4Cot²-3Tan²-3Cot²
4Tan²-3Tan²+4Cot²-3Cot²
Tan²-Cot²
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
700,353 = 50
133,533 = 500
596,967 = 500,000
632,295 = 5
Answer:
2$+(1.29$ x 30)=40.70$
Step-by-step explanation:
