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xenn [34]
2 years ago
15

I'm too lazy to do the work by myself anyway, someone please help me thank you <3 !!

Mathematics
1 answer:
In-s [12.5K]2 years ago
7 0

Answer:

98

Step-by-step explanation:

10-3=7

7x7=49

49 x2 = 98

Have a good day! uwu

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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
Find the equation of a line that passes through the points (4,0) and (2,4).
Ksju [112]

Answer:

answer is y=-2x+8 is the eq of given line

3 0
2 years ago
How long will it take a person to walk from Toronto to
storchak [24]

Answer:

About 947 hours, walking at 5 km (3 miles)/hour.

Walking 8 hours /day [at that pace you would need to be an Olympic Athlete to keep up the pace] it will take about four months.

Walk slower, it will take longer.

4 0
2 years ago
Write the equation of a quadratic function who has the vertex of (4,-7)
timama [110]

Given:

The vertex of a quadratic function is (4,-7).

To find:

The equation of the quadratic function.

Solution:

The vertex form of a quadratic function is:

y=a(x-h)^2+k          ...(i)

Where a is a constant and (h,k) is vertex.

The vertex is at point (4,-7).

Putting h=4 and k=-7 in (i), we get

y=a(x-4)^2+(-7)

y=a(x-4)^2-7

The required equation of the quadratic function is y=a(x-4)^2-7 where, a is a constant.

Putting a=1, we get

y=(1)(x-4)^2-7

y=(x^2-8x+16)-7             [\because (a-b)^2=a^2-2ab+b^2]

y=x^2-8x+9

Therefore, the required quadratic function is y=x^2-8x+9.

6 0
2 years ago
5.1x=-15.3 solve for x
LekaFEV [45]

Answer


x=-3


Hope it helps!!

3 0
3 years ago
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