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castortr0y [4]
3 years ago
14

How to find the midpoint of the line segment with given endpoint

Mathematics
1 answer:
Komok [63]3 years ago
5 0
X2 + x1 divided by 2. then y2 +y1 divided by 2. = midpoint, I think.
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horn y ?????????????????????? ??????? ????????????bgfbxfdcfyctzfxc hcyxydyxhc h cyxtxf y ffsdvbnbf vgrgbxxcb hhfvsd am dj di​
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Answer:

<em>n o  </em>?????????????????????? ??????? ????????????bgfbxfdcfyctzfxc hcyxydyxhc h cyxtxf y ffsdvbnbf vgrgbxxcb hhfvsd am dj di​

7 0
3 years ago
Read 2 more answers
70% of a number is 98.
Anni [7]
Set up the proportion:
70% - 98
90% - x

\frac{70\%}{90\%}=\frac{98}{x} \\&#10;\frac{7}{9}=\frac{98}{x} \\&#10;\frac{7}{9}x=98 \\&#10;x=98 \times \frac{9}{7} \\&#10;x=14 \times 9 \\&#10;x=126

The answer is D. 126.
8 0
3 years ago
Order the cities in the table from least to greatIest distance from sea level
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8 0
3 years ago
Find all solutions to the following system of equations -y²+6y+x-9=0 ; 6y=x+27.Illustrate with a graph.
Dimas [21]

Answer:

The solution to the system is the pair (9, 6)

Step-by-step explanation:

Hi!

First, let´s write the system of equations:

-y² + 6y + x -9 = 0

6y = x +27

The solutions of the system are the pairs (x, y) that satisfy both equations.

Let´s take the second equation and solve it for x:

6y = x +27

Subtract 27 from both sides of the equation

6y - 27 = x

Now, we can replace x in the first equation and solve it for y:

-y² + 6y + x -9 = 0

-y² + 6y + 6y - 27 -9 = 0

-y² + 12y - 36 = 0

Notice that -y² + 12y - 36 = -(y - 6)², then:

-(y - 6)² = 0

y - 6 = 0

y = 6

(alternatively you can solve the quadratic equation using the quadratic formula).

Now let´s find the value of x:

x = 6y -27

x = 6·6 -27

x = 9

The solution to the system is the pair (9, 6)

Please see the attached figure. The point where the curves intersect is the solution to the system.

6 0
3 years ago
Find the work required to move an object in the force field F = e^x + y (1, 1, z) along the straight line from A(0, 0, 0) to B(-
DENIUS [597]

\vec F(x,y,z)=(e^x+y)(1,1,z)

is conservative if we can find a scalar function f such that \nabla f=\vec F. This would require

f_x=e^x+y

f_y=e^x+y

f_z=(e^x+y)z

Integrating both sides of the first equation wrt x gives

f(x,y,z)=e^x+xy+g(y,z)

Differentiating both sides of this wrt y gives

f_y=x+g_y=1\implies g_y=1-x\implies g(y,z)=y-xy+h(z)

but we assumed g was a function of y and z, independent of x. So there is no such f and \vec F is not conservative.

To find the work, first parameterize the path (call it C) by

\vec r(t)=(1-t)(0,0,0)+t(-4,5,-5)=(-4t,5t,-5t)

for 0\le t\le1. Then

\vec r'(t)=(-4,5,-5)

and the work is given by the line integral,

W=\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1(e^{-4t}+5t,e^{-4t}+5t,-(e^{-4t}+5t)5t)\cdot(-4,5,-5)\,\mathrm dt

W=\displaystyle\int_0^1(125t^2+5t+(25t+1)e^{-4t})\,\mathrm dt=\boxed{\frac{2207}{48}-\frac{129}{16e^4}}

8 0
3 years ago
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