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3 years ago

Solve:

Mathematics

2 answers:

harina [27]3 years ago

8 0

Answer:

$x = 0$

Step-by-step explanation:

Write the number in exponential form with the base of 2:

${2}^{x - 3} = {2}^{ - 3}$

Since the bases are the same, set the exponents equals.

$x - 3 = - 3$

Cancel equal terms on both sides of the equation:

$x = 0$

dedylja [7]3 years ago

5 0

<h2>Given</h2>

$2 {}^{x - 3} = \frac{1}{8}$

To Find value of 'x'

<h3><u>Solution</u></h3>

<u> $2 {}^{x - 3} = \frac{1}{8} \\ \\ \implies2 {}^{x - 3} = 2 {}^{ - 3} \\ \\ \\ Since \: Bases \: Are \\ Equal \: We \: Need \\ To \: Evaluate \\ The \: Powers \\ \\ \implies \: x - 3 = - 3 \\ \\ \implies x = - 3 + 3 \\ \\ \implies \: x = 0$ </u>

<u> $\fbox \red{Hope This Helps You ❤}$ </u>

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ruslelena [56]

Answer:

(20,24) (17,20) and (8,8)

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Step-by-step explanation:

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3 years ago

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Maksim231197 [3]

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3 years ago

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Sever21 [200]

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6 0

3 years ago

3 to the 2 power times 5 to the 3 times 8 divided by 3 times 5 to the 3 times 8

IceJOKER [234]

Answer:

Step-by-step explanation:

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3 0

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