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solong [7]
4 years ago
8

Write a common denominator for 1/4 and 5/6

Mathematics
2 answers:
Verizon [17]4 years ago
7 0
The common denominator would be 12 .
You have to multiply 1/4 by 3/3 and 5/6 by 2/2 because if we want them to be out of 12 we have to multiply them by those numbers.
maxonik [38]4 years ago
3 0
A common denominator is 12
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A saleswoman received $40 per day, plus a commission of 2% on all sales over $300 per week. What was her salary for a 5-day week
ExtremeBDS [4]
2 % = 2/100 = 0.02 
So her salary is broken down into two parts. 
The systematic part is $40 per day for 5 days, so 40*5. 
The changing part is 2% of weekly sales over 300, so 0.02(1260-300). 

Salary = 5*40 +0.02(1260-300)=$219.20
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How to find 8x4 by using Distributive property
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Quartile 1 is 211. quartile 2 is 226 , and quartile 3 is 265.5
 
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3 years ago
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Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
How do you end up with 0=0
Eva8 [605]
Ok so first plug in y for in the equation 2x-2y=12
Therefore 2x - 2(x -6)=12
Then distribute the two to the x and the -6 
2x-2x+12=12
The 2x's cancel leaving you with 
12=12
From there you could subtract the 12 from both sides to get 0=0, I assume
4 0
3 years ago
Read 2 more answers
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