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amid [387]
3 years ago
12

Solve by the matrix method , x+y/5 -1=y-x​

Mathematics
1 answer:
PIT_PIT [208]3 years ago
6 0

10x - 4y = 5 (if it requires shorten)

Step-by-step explanation:

x + y/5 - 1 = y - x

<=> (5x + y - 1.5)/5 = 5(y - x)/5

<=> 5x + y - 5 = 5y - 5x

<=> 5x + 5x + y - 5y = 5

<=> 10x - 4y = 5

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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}}

8 0
3 years ago
Suppose point j is the midpoint of segment lm where j(5, -1) and m(0, 6). what are the coordinates of end point l?
zalisa [80]

If the midpoint of segment lm where j(5, -1) and m(0, 6), then the coordinates of end point l is (10,-8)..

Midpoint:

Midpoint of a line segment is the point that is halfway between the endpoints of the line segment.

The formula for the midpoint is,

(x_m,y_m)=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

where,

(x_m,y_m) = coordinates of the midpoint

(x₁, y₁)   = coordinates of the first point

(x₂, y₂) = coordinates of the second point

Given,

Midpoint = j = (5,-1)

Second point =  m = (0,6)

Now, we have to find the first midpoint l.

Let (x,y) be the coordinates of l.

Then apply the values on the formula,

Then we get,

(5,-1)=(\frac{x+0}{2},\frac{y+6}{2})

Compare the values in order to find the coordinates,

=> 5 = (x + 0) / 2

=> 5 x 2 = x + 0

=> 10 = x

Similarly,

=> -1 = (y + 6) / 2

=> -1 x 2 = y + 6

=> -2 = y + 6

=> -2 - 6 = y

=> y = -8

Therefore, the coordinates of l is (10,-8).

To know more about Midpoint here.

brainly.com/question/28224145

#SPJ4

3 0
1 year ago
A student playing a computer chess game gets 5 points every time he wins a round. The computer gets 3 points every time it wins
Kipish [7]
I am sorry but i don't know, But the student won 32 times. I did the math on paper and then on my calculator. I am really sorry i couldn't help you. 'ω' Hope you do good on this question and again sorry i could not help you.<span />
4 0
3 years ago
Is the relation a function? Why or why not?<br><br> {(–3, –2), (–1, 0), (1, 0), (5, –2)}
tankabanditka [31]

Answer:

yes, the given relation is a function.

Step-by-step explanation:

The given relation is

{(–3, –2), (–1, 0), (1, 0), (5, –2)}

A relation is called function if each element of the domain is paired with exactly one element of the range.

It means for each value of x there exist a unique value of y.

In the given relation for each value of x there exist a unique value of y.

Therefore the required solution is yes and this relation is a function.

5 0
3 years ago
What is the solution got the graphed system of equations ? <br><br> Help please :(
Anvisha [2.4K]

Answer:

It's where they meet! I can't really see the coordinate clearly, but if I could I would tell you already.

8 0
3 years ago
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