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Hoochie [10]
3 years ago
8

I need help with this

Mathematics
1 answer:
AleksAgata [21]3 years ago
5 0
I don't see the problem sorry
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The value of y varies directly as x and y is 9 when x is -3 what constant of variation k?
Lady bird [3.3K]
Y/x=k

when y=9, then x= -3
        9/-3=-3
⇒ k= -3
7 0
3 years ago
PLEASE HELP ME EVERYONE HAS GOTTEN THIS WRONG ITS GONNA SUBMIT IN 15 MIN PLS HELP THANK YOU
LiRa [457]

Answer:

x=4.8

This here you use Pythagorean Theorem.

Step-by-step explanation:

a^2+b^2=c^2

The longest side is c. And the two legs are a and b.

So it'd be 11^2+x^2=12^2

121+x^2=144

Subtract 121 from both sides.

x^2=23

Then take the √ of x^2 and 23

So √x^2=√23

x=4.8

Hope this helps

5 0
3 years ago
Q. #3 I need help please
Leto [7]
You need to multiply 43x43
7 0
4 years ago
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A planet has a circular orbit around a star. It is a distance of 65,000,000 km from the centre of the star. It orbits at an aver
guapka [62]

Answer:

Step-by-step explanation:

Orbit = 2*pi * r

r = 65000000 km

Orbit = 65000000 * 2 * 3.14

Orbit = 406 200 000             km

d = 406 200 000

s = speed = 65000 km / hr

t = time = d / s

t = 406,000,000 / 65000

t = 6280 hours.

1 day = 24 hours.

t = 6280 / 24 = 261.67 days

8 0
3 years ago
Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

Now, let us compute the value using Green's theorem.

According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2

3 0
3 years ago
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