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3 years ago

Describe the solution of the equation x^2+7x+2=0

1 answer:

4 0

This is a quadratic equation with a general equation of ax^2 + bx + c.
The quadratic formula can help to get the roots of the equation. We know the highest degree of that equation is 2; so there will be also two roots.
The quadratic formula is
x = [-b ± √(b^2 - 4ac)] / 2a

With a = 1, b = 7, c = 2,
x = {-7 ± √[(7)^2 - 4(1)(2)]} / 2(1) = (-7 ± √41) / 2

So the two roots are
x1 = (-7 + √41) / 2 = -0.2984
x2 = (-7 - √41) / 2 = 0.2984

This is also another way of factorizing the equation
(x + 0.2984)(x + 0.2984) = x^2 + 7x + 2

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marusya05 [52]

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6 0

3 years ago

Read 2 more answers

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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5 0

2 years ago

Thank you:)))))))))))))

Anastasy [175]

Your welcomeeeeeeeeeee

3 0

3 years ago