Answer:
2
Step-by-step explanation:
OMG....this one took me hours to solve...I'm guess 2?
The answer is 426/5 and in mixed form is 85 1/5
Answer:
Solution given:
A(a,b) lie in 6x-y =1
substituting value of A in equation ,we get
6a-b=1...................(1)
B(b,a) lie in 2x - 5y = 5
substituting value of B in equation ,we get
2b-5a=5
b=5(a+1)/2............(2)
substituting value of b in equation 1 ,we get
6a-5(a+1)/2=1
12a-5a-5=2
7a=2+5
a=
=1
substituting value of a in equation 1 ,we get
6×1-b=1
6-1=b
b=5
<u>so</u><u> </u><u>points</u><u> </u><u>are</u><u> </u><u>A</u><u>(</u><u>1</u><u>,</u><u>5</u><u>)</u><u> </u><u>and</u><u> </u><u>B</u><u>(</u><u>5</u><u>,</u><u>1</u><u>)</u>
we have
equation of line having two points
![(y - y1) = \frac{y2 - y1}{x2 - x1} (x - x1)](https://tex.z-dn.net/?f=%28y%20-%20y1%29%20%3D%20%20%5Cfrac%7By2%20-%20y1%7D%7Bx2%20-%20x1%7D%20%28x%20-%20x1%29%20%20)
![(y - 5) = \frac{1-5}{5-1} (x - 1)](https://tex.z-dn.net/?f=%28y%20-%205%29%20%3D%20%20%5Cfrac%7B1-5%7D%7B5-1%7D%20%28x%20-%201%29%20%20)
![(y - 5) = -1 (x - 1)](https://tex.z-dn.net/?f=%28y%20-%205%29%20%3D%20-1%20%28x%20-%201%29%20%20)
y-5=1-x
<u>x+y=6 is a required equation of line AB.</u>
Answer: " y = 21 " .
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Cross multiply:
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(2y)(4) = 7(y + 3) ;
8y = 7y + 21
Subtract 7y from each side of the equation;
8y - 7y = 7y + 21 - 7y ;
y = 21 .
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Answer:
All the properties of a rhombus apply (the ones that matter here are parallel sides, diagonals are perpendicular bisectors of each other, and diagonals bisect the angles). All the properties of a rectangle apply (the only one that matters here is diagonals are congruent). All sides are congruent by definition.