Match each polynomial with its degree
1 answer:
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Answer:
The lower supports are <u>Congruent Rectangles</u> and the area of the two supports is<u> 24 square meters.</u>
The upper arch can be decomposed as one semicircle with radius<u> 6 </u>meters minus a semicircle with radius 3 meters.
The area of the archway is <u>(13.5 π</u> + 24) square meters.
Step-by-step explanation:
See attachment for the firgure,
In order to determine the area of archway = The area of upper support + area of lower support.
As given, the lower support are two congruent rectangles consists of dimension 3 m × 4 m
Therefore, the area of lower support can be written as,
The area of lower support = 3 × 4 + 3 × 4 = 12 + 12 = 24 square m.
Now, the upper support arch can be decomposed as the two concentric semi circles having radius 3 m and 6 m,
Hence, the area of the upper support = Area of semi circle having radius i.e 6 m - Area of semi circle having radius i.e 3 m
=> π6²/2 - π3²/2= 27π/2= 13.5π square m
Therefore, the area of the archway = (13.5 π + 24) square meters
Answer:
<u>y = -x² + 4</u>
Step-by-step explanation:
The equation of the parabola in the vertex form is:
y = a (x-h)² + k
Where: (h,k) the coordinates of the vertex & a is a multiplier
The parabola has a vertex at ( 0,4 )
So, h = 0 , k = 4
∴ y = a (x-0)² + 4
∴ y = a x² + 4
The parabola passes through points ( 2,0 )
∴ 0 = a 2² + 4
∴ 4 a = -4 ⇒ a = -4/4 = -1
∴ y = -x² + 4
So, the equation of a parabola that has a vertex at ( 0,4 ) and passes through points ( 2,0 ) is <u>y = -x² + 4</u>
See the attached figure.
