Answer:
There are 2
Step-by-step explanation:
Answer:
The area of the square adjacent to the third side of the triangle is 11 units²
Step-by-step explanation:
We are given the area of two squares, one being 33 units² the other 44 units². A square is present with all sides being equal, and hence the length of the square present with an area of 33 units² say, should be x² = 33 - if x = the length of one side. Let's make it so that this side belongs to the side of the triangle, to our convenience,
x² = 33,
x = 
.... this is the length of the square, but also a leg of the triangle. Let's calculate the length of the square present with an area of 44 units². This would also be the hypotenuse of the triangle.
x² = 44,
x = 
.... applying pythagorean theorem we should receive the length of a side of the unknown square area. By taking this length to the power of two, we can calculate the square's area, and hence get our solution.
Let x = the length of the side of the unknown square's area -
= 
+ 
,
x = 
... And squared is 11, making the area of this square 11 units².
Answer:
120
Step-by-step explanation:
The LCM of 24 and 30 is 120. To find the least common multiple (LCM) of 24 and 30, we need to find the multiples of 24 and 30 (multiples of 24 = 24, 48, 72, 96 . . . 120; multiples of 30 = 30, 60, 90, 120) and choose the smallest multiple that is exactly divisible by 24 and 30, i.e., 120.
Answer: C) No solution
Let’s solve this. First we need to put the first equation into y=mx+b form. Let’s do this now.
First equation-
2•4y=10
*subtract 2*
4y=8
*divide 4*
y=2
Now that we know the value of y, let’s substitute that into the second equation.
2•4y= -10
2•4(2)= -10
2•8= -10
16= -10
When solved, these equations produce a result where both sides do not equal each other. 16 does not equal -10, making this have no solution.
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