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Viktor [21]
3 years ago
6

Which ordered pair can be removed so that the

Mathematics
2 answers:
anzhelika [568]3 years ago
6 0

Answer:

B

Step-by-step explanation:

nalin [4]3 years ago
4 0
Answer:
(1, 3) can be removed so that the resulting graph represents a function.

Explanation:
A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y.
Here,
X = {-5, -4, -2, 1, 2, 5}
Y = {-4, -3, -2, 1, 2, 3}
Relation from X to Y : {(-5, -3), (-4, -4), (-2, 2), (1, -2), (1, 3), (2, 1), (5, -4)}
This relation is not a function from X to Y because the element 1 in X is related to two different elements, -2 and 3
So, we have to remove either (1, -2) or (1, 3) in order to make this a function.
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Mr.Jenkins has 48 paintbrushes and 60 tubes of paint to put into cups.The greatest common factor for the number of paint brushes
Flauer [41]

Answer:

Mr. Jenkins has 12 cups.

Step-by-step explanation:

Consider the provided information.

Mr.Jenkins has 48 paintbrushes and 60 tubes of paint to put into cups.

It is given that The greatest common factor for the number of paint brushes and the number of tubes of paint is equal to the number of cups Mr.Jenkins has.

First find the greatest common factor of 48 and 60.

48 = 2×2×2×2×3

60 = 2×2×3×5

Hence, the greatest common factor is 2×2×3=12

Therefore, Mr. Jenkins has 12 cups.

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Use stokes' theorem to evaluate c f · dr where c is oriented counterclockwise as viewed from above. f(x, y, z = xyi + 5zj + 7yk,
Helga [31]
The intersection can be parameterized by

C:=\mathbf r(t)=\begin{cases}x(t)=6\cos t\\y(t)=6\sin t\\z(t)=5-6\cos t\end{cases}

with 0\le t.

By Stoke's theorem, the integral of \mathbf f(x,y,z)=xy\,\mathbf i+5z\,\mathbf j+7y\,\mathbf k along C is equivalent to

\displaystyle\int_C\mathbf f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r(t)=\iint_S\nabla\times\mathbf f\,\mathrm dS

where S is the region bounded by C. The line integral reduces to

\displaystyle\int_0^{2\pi}(36\sin t\cos t\,\mathbf i+(25-30\cos t)\,\mathbf j+42\sin t\,\mathbf k)\cdot(-6\sin t\,\mathbf i+6\cos t\,\mathbf j+6\sin t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{2\pi}(54(\cos3t-\cos t)-30(3\cos2t-5\cos t+3)+(126-126\cos2t)\,\mathrm dt
=\displaystyle\int_0^{2\pi}(36+96\cos t-216\cos2t+54\cos3t)\,\mathrm dt
=72\pi
4 0
4 years ago
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