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3 years ago

Given a rectangle with lengths of 7 meters and widths of 5 metres,what would be the perimeter

Mathematics

1 answer:

Vsevolod [243]3 years ago

3 0

Answer:

Perimeter = 24 meters

step by step explanation

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ILL BRAINLIEST YOU PLEASE HELP ME

Degger [83]

Answer:

63 cause there equal

Step-by-step explanation:

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3 years ago

Find the equation of the quadratic function whose graph is a parabola containing the points (0,2), (3,26), and (-1,6).

Arlecino [84]

Answer: y = 3x^2 -x + 2

Step-by-step explanation:

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3 years ago

Carson has two cans of red paint. The first can contains 3 pints of red paint. The second can contains 4 16 pints of red paint.

allsm [11]

Answer: 1⁵/₈ pints

Step-by-step explanation:

First can contained 3¹/₄ pints and second can contained 6¹/₂.

To make them equal, first find the different;

= 6¹/₂ - 3¹/₄

= 6.5 - 3.25

= 3.25

Divide this difference by 2;

= 3.25/2

= 1.625

= 1⁶²⁵/₁₀₀₀

= 1⁵/₈ pints

If Carson pours 1⁵/₈ pints from the second Can, it would increase the first can to 4⁷/₈ pints whilst reducing the second can to 4⁷/₈ pints as well.

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3 years ago

How do I do it? I’m get confused what do I type in the box?

erastovalidia [21]

I think that’s called (translations)if i’m not wrong

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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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2 years ago