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Sergeeva-Olga [200]
3 years ago
10

((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2

Mathematics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

  256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

  (2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

  (a^b)^c = a^(b·c)

  (a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

  (2^2)(2^((-3)(-2)) = (2^2)(2^6)

  = 2^(2+6)

  = 2^8 = 256

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What survival rate in the 10th year would make the average rate of loss over the 10-year period equal to 38.0%?
Morgarella [4.7K]

Answer: 63%

Step-by-step explanation:

First find the rate of loss that would cause the average rate of loss over the 10-year period equal to 38.0%.

Assume that rate is x.

38 = (36.2 + 29.0 + 46.2 + 37.5 + 40.9 + 40.0 + 32.6 + 40.5 + 40.1 + x) / 10

38 = (343 + x ) / 10

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x = 380 - 343

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2 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}) dxis 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}) dxis 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
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Y-1/4=-3/8 solve for y
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Answer:

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Step-by-step explanation:

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A translation to the left means subtract 2 from the x- coordinate

A translation down means subtract 4 from the y- coordinate

Then

(8, 4 ) → (8 - 2, 4 - 4 ) → (6, 0 )

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2 years ago
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