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olganol [36]
3 years ago
10

The vertices of a parallelogram are A(x1, y1), B(x2, y2), C(x3, y3), and D(x4, y4). Which of the following must be true if paral

lelogram ABCD is proven to be a rectangle? A. and B. and C. and D. and

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
8 0

Answer:

The following points are not arranged in a parallelogram or rectangle order.

Step-by-step explanation:

Well first we need to graph the following.

A(1,1) B(2,2) C(3,3) D(4,4)

By looking at the image below we can tell it is not any shape, it’s not a parallelogram or a rectangle.

It is a line with a slope of 1 or x.

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Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

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3 years ago
Finding the Slope: Identify the slope. Will give 5 points if correct​
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3 years ago
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Number two should correspond for the graphs, with graph paper, just follow the color coded lines. Hope this helps.

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The sum of two numbers is "-1" One number is 12 more than the other one. Find the numbers.
ruslelena [56]

Answer:

x = 5.5

y = - 6.5

Step-by-step explanation:

Let one of the numbers = x

Let the other number = y

x + y = - 1

x = y + 12    Put the second equation into the first one

y + 12 + y = - 1       Subtract 12 from both sides

y + y = - 1 - 12        Combine both left and right sides

2y = - 13                Divide by 2

2y/2 = - 13/1

y = - 6.5

x + y = - 1

x - 6.5 = - 1            Add 6.5 to both sides

x = 5.5

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3 years ago
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