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lesya692 [45]
3 years ago
5

The road map indicates that it is 10 miles from Vacaville to Fairfield. From the information on the road map, it follows that Va

caville and Fairfield really are 10 miles apart. what type of argument?
Mathematics
1 answer:
slamgirl [31]3 years ago
3 0

Answer: Inductive argument.

Step-by-step explanation:

  • An argument can have one or more premises but there is only one conclusion to it.

The arguments are of two types : Inductive  (uses pattern or signs to get a conclusion ) and deductive (Uses general facts or defines or theory to decide any conclusion)

The given argument : The road map indicates that it is 10 miles from Vacaville to Fairfield. From the information on the road map, it follows that Vacaville and Fairfield really are 10 miles apart.

The given argument is the argument that is based on signs (maps are signs) which comes under inductive arguments.

Thus, the given argument is an inductive argument.

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Answer:

<em>jzjddndndndndhzzhxh</em>

Step-by-step explanation:

hggahgababsbsj

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2 years ago
Prove this identity sin(2A)=2sinAcosA.
34kurt
The solution for proving the identity is as follows:

sin(2A) = sin(A + A) 
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<span>Therefore, sin(2A) = 2sinAcosA

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7 0
2 years ago
TIMED TEST PLEASE HELP
xxMikexx [17]
1 ) first offer
total payments
375.76×12×4
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18,036.48−16,000
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6 0
3 years ago
Read 2 more answers
Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ
liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

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Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

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We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2

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2 years ago
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