Answer:
64
Step-by-step explanation:
Your welcome
The longest possible integer length of the third side of the triangle is 6 < x < 28
The sum of any two sides must be greater than the third side for a triangle to exist
let the third side be x
x + 11 > 17 and x + 17 > 11 and 11 + 17 > x
x > 6 and x > - 6 and x < 28
The longest possible integer length of the third side of the triangle is 6 < x < 28
The length of the 3 sides of a triangle needs to always be among (however no longer the same) the sum and the difference of the opposite two sides. As an example, take the instance of two, 6, and seven. and. consequently, the third side period should be extra than 4 and less than 8.
Learn more about triangles here: brainly.com/question/1675117
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5 1/4 in decimal form is 5.25 because 25 is 1/4 of 100.
*see attachment for the figure described
Answer:
5 units
Step-by-step explanation:
==>Given the figure attached below, let where FH and EG intercepted be K.
Since FH are midpoints of parallel lines, KE = KG = x.
Given that the area of the kite EFGH = 35 square units, and we know the length of one of the diagonals = HF = KF + KH = 2 + 5 = 7, we can solve for x using the formula for the area of a kite.
Area of kite = ½ × d1 × d2
Where d1 = KH = 7
d2 = EG = KE + KG = x + x = 2x
Area of kite EFGH = 35
THUS:
35 = ½ × 7 × 2x
35 = 1 × 7 × x
35 = 7x
Divide both sides by 7
35/7 = x
x = 5
Answer:
Aid paid for 5115
You pay 1705
Step-by-step explanation:
6820×.75= 5115
6820-5115= 1705