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:
D) 2x² + 4x - 30
Step-by-step explanation:
Follow the FOIL method:
FOIL =
First
Outside
Inside
Last
...and is the order you multiply to solve:
(x + 5)(2x - 6) =
(x)(2x) = 2x²
(x)(-6) = -6x
(5)(2x) = 10x
(5)(-6) = -30
Combine like terms. Simplify:
2x² -6x + 10x - 30 = 2x² (-6x + 10x) - 30 = 2x² + 4x - 30
2x² + 4x - 30 is your answer.
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Answer:
58
Step-by-step explanation:
the formula is -4n+206
Willard F. Libby used Geiger M<span>üller tubes to date artifacts with known ages. </span>
Answer:
H
Step-by-step explanation:
add the numbers in the brackets first and