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Leto [7]
3 years ago
8

The force F of attraction between two bodies varies jointly as the weights of the two bodies and inversely as the square of the

distance d between them. Express this fact as a variation using c as a constant. Use M1 and M2 for the weights of the two bodies.
Mathematics
1 answer:
Maksim231197 [3]3 years ago
8 0
Since we are to express the given conditions as a joint variation, we employ the use of k as our constant proportionality which we will denote in this item as k. 
 From the sentence,
                            F = kM₁M₂/d²
as we can see those in the denominator as those that vary inversely to the given parameter. 
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svetoff [14.1K]

The answer is F because a common point is (0,-4)

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3 years ago
Solve the following equation simultaneously 2a-3b+10=0, 10a+6b=8​
tia_tia [17]

2a -3b +10 =0\\\\10a+ 6b -8 =0\\\\\text{Using cross multiplication method}\\\\\\ \dfrac a{(-3)(-8)  - 6(10)} = \dfrac b{10(10) - 2(-8)} = \dfrac 1{ 2(6) - (-3)(10)}\\\\\\\implies  \dfrac a{24-60} = \dfrac b{100+16} = \dfrac 1{12+30}\\\\\\\\\implies -\dfrac {a}{36} = \dfrac b{116} = \dfrac 1{42}\\\\\\\text{Hence}\\\\a = - \dfrac{36}{42} = -\dfrac 67 \\\\\\ b= \dfrac{116}{42} = \dfrac{58}{21}

3 0
3 years ago
The coordinates of the three vertices of the triangle are A (x1, y1), B (x2, y2), C (x3, y3).
kaheart [24]

Answer:

Step-by-step explanation:

Using the section formula, if a point (x,y) divides the line joining the points (x  

1

​

,y  

1

​

) and (x  

2

​

,y  

2

​

) in the ratio m:n, then  

(x,y)=(  

m+n

mx  

2

​

+nx  

1

​

 

​

,  

m+n

my  

2

​

+ny  

1

​

 

​

)

The vertices of the triangle are given to be (x  

1

​

,y  

1

​

),(x  

2

​

,y  

2

​

) and (x  

3

​

,y  

3

​

). Let these vertices be A,B and C respectively.  

Then the coordinates of the point P that divides AB in l:k will be

(  

l+k

lx  

2

​

+kx  

1

​

 

​

,  

l+k

ly  

2

​

+ky  

1

​

 

​

)

The coordinates of point which divides PC in m:k+l will be  

⎩

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎧

​

 

m+k+l

mx  

3

​

+(k+l)  

(l+k)

lx  

2

​

+kx  

1

​

 

​

 

​

,  

m+k+l

my  

3

​

+(k+l)  

(l+k)

ly  

2

​

+ky  

1

​

 

​

 

​

 

⎭

⎪

⎪

⎪

⎬

⎪

⎪

⎪

⎫

​

 

⇒(  

m+k+l

kx  

1

​

+lx  

2

​

+mx  

3

​

 

​

,  

m+k+l

ky  

1

​

+ly  

2

​

+my  

3

​

 

​

)

6 0
3 years ago
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Serhud [2]

Answer:

10.5

Step-by-step explanation:

7 0
3 years ago
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Please help!! 15 points for the answer.
Anarel [89]
A) (5s + 2 / 4) x (4) = Perimeter
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C) 22 + 48 = 70


4 0
3 years ago
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