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3 years ago

In the first pic ABC=DBE find value of x, ABC=DEF find F in the second one

Mathematics

2 answers:

Ivenika [448]3 years ago

5 0

if ABC = DBE

then <a = <d and <c = <e

<c = <e

<c = 83

<a +<b + <c = 180 triangles add to 180

x+8 + 51+ 83 = 180

combine like terms

x+142 = 180

subtract 142 from each side

x = 38

ABC = DEF

then <a = <d <b = <e and <c = <f

125 = <e

<d+<e+ <f = 180 triangles add to 180

35+125+ <f = 180

combine like terms

160 +<f = 180

subtract 160 from each side

<f = 120

FinnZ [79.3K]3 years ago

3 0

First problem:

x + 8 + 51 + 83 = 180

x + 142 = 180

x = 38

Second problem:

m<C + 35 + 125 = 180

m<C + 160 = 180

m<C = 20

m<F = m<C = 20

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$\left(x-8\right)^2+16=x^2-16x+80$

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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

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⇒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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