Answer:
48 x 2 if its to and from use caculator and do " 48 x 2"
Step-by-step explanation:
sorry if im wrong
Answer:
x=5.657 y=5.657
Step-by-step explanation:
<h3>☂︎ Answer :- </h3>
<h3>☂︎ Solution :- </h3>
- LCM of 5 , 18 , 25 and 27 = 2 × 3³ × 5²
- 2 and 3 have odd powers . To get a perfect square, we need to make the powers of 2 and 3 even . The powers of 5 is already even .
In other words , the LCM of 5 , 18 , 25 and 27 can be made a perfect square if it is multiplied by 2 × 3 .
The least perfect square greater that the LCM ,
☞︎︎︎ 2 × 3³ × 5² × 2 × 3
☞︎︎︎ 2² × 3⁴ × 5²
☞︎︎︎ 4 × 81 × 85
☞︎︎︎ 100 × 81
☞︎︎︎ 8100
8100 is the least perfect square which is exactly divisible by each of the numbers 5 , 18 , 25 , 27 .
Answer:
2
Step-by-step explanation:
2 is a value of x that would not make the set a function because a function can't have multiple coordinates wit hthe same domain (x-coordinate). 2 is a domain that has already been used in the point (2,7). If this is used in (x,5), the point would become (2,5) and repeat a domain.
Note:
The x-coordinate could also be 1 or 6 because they have been used as domains as well.
Hope it helps!
Intersection of the first two lines:
Multiply the first equation by 4 and the second by 5:
Subtract the two equations:
Plug this value for y in one of the equation, for example the first:
So, the first point of intersection is 
We can find the intersection of the other two lines in the same way: we start with
Use the fact that x and y are the same to rewrite the second equation as
And since x and y are the same, the second point is 
So, we're looking for a line passing through and . We may use the formula to find the equation of a line knowing two of its points, but in this case it is very clear that both points have the same coordinates, so the line must be 
In the attached figure, line 
is light green, line 
is dark green, and their intersection is point A.
Simiarly, line 
is red, line 
is orange, and their intersection is B.
As you can see, the line connecting A and B is the red line itself.
