A right triangle's longest side is the hypotenuse
let x=longest, y=middle, and z=shortest
x=y+2
y=2z-1
therefore x=(2z-1)+2=2z+1
find z
z^2+y^2=x^2 by Pythagorean theorem
plug in x and y in terms of z
z^2+(2z-1)^2=(2z+1)^2
z^2+4z^2-4z+1=4z^2+4z+1
subtract the right-hand side's value from the left-hand side's
z^2-8z=0
z(z-8)=0
z=0, 8
z cannot be zero as the sides must have some value to it.
Therefore the shortest side is equal to 8
Part A
<h3>Answer:
h^2 + 4h</h3>
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Explanation:
We multiply the length and height to get the area
area = (length)*(height)
area = (h+4)*(h)
area = h(h+4)
area = h^2 + 4h .... apply the distributive property
The units for the area are in square inches.
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Part B
<h3>Answer:
h^2 + 16h + 60</h3>
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Explanation:
If we add a 3 inch frame along the border, then we're adding two copies of 3 inches along the bottom side. The h+4 along the bottom updates to h+4+3+3 = h+10 along the bottom.
Similarly, along the vertical side we'd have the h go to h+3+3 = h+6
The old rectangle that was h by h+4 is now h+6 by h+10
Multiply these expressions to find the area
area = length*width
area = (h+6)(h+10)
area = x(h+10) ..... replace h+6 with x
area = xh + 10x .... distribute
area = h( x ) + 10( x )
area = h( h+6 ) + 10( h+6 ) .... plug in x = h+6
area = h^2+6h + 10h+60 .... distribute again twice more
area = h^2 + 16h + 60
You can also use the box method or the FOIL rule as alternative routes to find the area.
The units for the area are in square inches.
If x=n then it would have 2 solutions because it would be x^2 = c which has a positive and negative solution but idk if that is what you are asking for
Answer:
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Step-by-step explanation:
Given
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← divide numerator/ denominator by 14 )
= -
