Would 45% of 160 be greater than or less than 80?
1 answer:
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Answer:
5 units
Step-by-step explanation:
Let point O be the point of intersection of the kite diagonals.
|OF| = 2, |OH| = 5
|FH| = |OF| + |OH| = 2 + 5 = 7
FH and EG are the diagonals of the kite. Hence the area of thee kite is:
Area of kite EFGH = (FH * EG) / 2
Substituting:
35 = (7 * |EG|) / 2
|EG| * 7 = 70
|EG| = 10 units
The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units
x = |GO| = |EO| = 5 units
A ) ![\sqrt[4]{ x^{3} } = x^{3/4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20x%5E%7B3%7D%20%7D%20%3D%20x%5E%7B3%2F4%7D%20)
B )
C )![\sqrt[10]{ x^{5} * x^{4}* x^{2} } = \sqrt[10]{ x^{11} }= x^{11/10}](https://tex.z-dn.net/?f=%20%5Csqrt%5B10%5D%7B%20x%5E%7B5%7D%20%2A%20x%5E%7B4%7D%2A%20x%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%5B10%5D%7B%20x%5E%7B11%7D%20%7D%3D%20x%5E%7B11%2F10%7D%20%20)
D )
Therefore: B = D
B )
C )
D )
Therefore: B = D