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antoniya [11.8K]
3 years ago
15

Write 98 as the product of prime factors in ascending order

Mathematics
1 answer:
Dimas [21]3 years ago
7 0

                98

       2           49

               7            7

 so the answer is 2 x 7 x 7, which is 2 x 7^2

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What reason should kamal add to all of the question marks in order to completer the proof
brilliants [131]

Answer: The answers is alternate interior angles.


Step-by-step explanation:  First of all, the questions marks given in the figure are renamed in the attached figure as (a), (b), (c) and (d).

For (a): Since AC is parallel to A'C' and A'D is a transversal for these two parallel lines, so, ∠CDB' = ∠B'A'C', because these are alternate interior angles.

For (b): Since BC is parallel to B'C' and A'B' is a transversal, so ∠BEB' = ∠A'B'C', because these are alternate interior angles.

For (c): Since AB is parallel to A'B' and AD is a transversal, so ∠BAC = ∠CDB', because these are alternate interior angles.

For (d): Since AB is parallel to A'B' and BE is a transversal, so ∠ABC = ∠BEB', because these are alternate interior angles.

Thus, all the questions marks are the reasons that the given angles are equal because they are alternate interior angles.

5 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}) 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

#SPJ4

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2 years ago
Pls pls pls help me this is due soon
Thepotemich [5.8K]

Answer:

x+18

Step-by-step explanation:

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Solve for x: −2(x + 3) = −2(x + 1) − 4
mash [69]
-2(x + 3) = -2(x + 1) - 4
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TIME LIMITED! PLEASE HELP!
Vlad1618 [11]

Answer:

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Step-by-step explanation:

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