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3 years ago

Solve the equation. x² + 6x+15=0

Mathematics

1 answer:

lara [203]3 years ago

3 0

Answer:

x = -3 + i√6 and x = -3 - i√6

Step-by-step explanation:

Let's apply the "completing the square" method to find the roots of this equation.

Take half of the coefficient 6 of x, square it and add this result to x^2 + 6x. Then subtract the same quantity:

x^2 + 6x + 15 becomes

x^2 + 6x + 3^2 - 3^2 + 15 = 0

Rewriting the first three terms as the square of a binomial, we ge:

(x + 3)^2 - 9 + 15 = 0, which simplifies to:

(x + 3)^2 + 6 = 0, or (x + 3)^2 = -6

Taking the square root of both sides:

x + 3 = ±i√6

Then the two roots are complex:

x = -3 + i√6 and x = -3 - i√6

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Elanso [62]

Answer:

a) 1/800 or 0.00125

b) i) 0.0013

ii) 0.001

c) 60%

Step-by-step explanation:

T = [tan(2×30)+1][2cos(30)-1] ÷ (y²-x²)

T = (tan60 + 1)(2cos30 - 1) ÷ (41² - 9²)

T = (sqrt(3) + 1)(sqrt(3) - 1) ÷ 1600

T = (3-1)/1600

T = 2/1600

T = 1/800

T = 0.00125

Error: 0.002 - 0.00125

0.00075

%error

0.00075/0.00125 × 100

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3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Given an integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ .

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑f(a+iΔx)Δx ,here Δx=(b-a)/n

$\int\limits^5_1 {x/(2+x^{3}) } \, dx$ =f(x)= $x/2+x^{3}$

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)= $[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}$

$\lim_{n \to \infty}$ ∑f(a+iΔx)Δx=

$\lim_{n \to \infty}$ ∑ $n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n$

=4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$

Hence the expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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