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statuscvo [17]
3 years ago
5

What is the distance between points A(7,3) and B(5,-1)

Mathematics
1 answer:
Soloha48 [4]3 years ago
3 0
We need to use  distance formula
Distance=√((x2-x1)²+(y2-y1)²)
Distance=√((7-5)²+(3-(-1))²)=√(4+16)=√20=2√5≈4.47
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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
Solve for k:<br>1/3k + 1/2 = 1/4k
son4ous [18]
\frac{1}{3}k+\frac{1}{2}=\frac{1}{4}k\ \ \ \ \ \ |\cdot12(to\ eliminate\ fractions)\\\\\frac{12}{3}k+\frac{12}{2}=\frac{12}{4}k\\\\4k+6=3k\\\\4k-3k=-6\\\\k=-6
3 0
3 years ago
ILL BRAINLIEST YOU PLEASE HELP ME
-Dominant- [34]

Answer:

x = 7/\sqrt{2} or 4.949747468

Step-by-step explanation:

45-45-90 triangles have a specific pattern.

The smaller sides are x, & the hypotenuse is x*\sqrt{2}

With this knowledge, we know we have to divide the hypotenuse by \sqrt{2} to get the smaller sides, so we divide 7 by the \sqrt{2} to find x.

4 0
3 years ago
The question <br> Najheheggebbsbv nhsygenbs
dezoksy [38]

Answer: 1.) √121= 11

Step-by-step explanation:

11•11=121

4 0
3 years ago
Suppose a shoe factory produces both low-grade and high-grade shoes. The factory produces at least twice as many low-grade as hi
umka2103 [35]

Answer:

The factory should produce 166 pairs of high-grade shoes and 364 pairs of low-grade shoes for maximum profit

Step-by-step explanation:

The given parameters for the shoe production are;

The number of low grade shoes the factory produces ≥ 2 × The number of high-grade shoes produced by the factory

The maximum number of shoes the factory can produce = 500 pairs of shoes

The number of high-grade shoes the dealer calls for daily ≥ 100 pairs

The profit made per pair of high-grade shoed = Birr 2.00

The profit made per of low-grade shoes = Birr 1.00

Let 'H', represent the number of high grade shoes the factory produces and let 'L' represent he number of low-grade shoes the factory produces, we have;

L ≥ 2·H...(1)

L + H ≤ 500...(2)

H ≥ 100...(3)

Total profit, P = 2·H + L

From inequalities (1) and (2), we have;

3·H ≤ 500

H ≤ 500/3 ≈ 166

The maximum number of high-grade shoes that can be produced, H ≤ 166

Therefore, for maximum profit, the factory should produce the maximum number of high-grade shoe pairs, H = 166 pairs

The number of pairs of low grade shoes the factory should produce, L = 500 - 166 = 334 pairs

The maximum profit, P = 2 × 166 + 1 × 364 = 696

6 1
3 years ago
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