I did the calculations of A(t)=50(0.8)5
and i got 200 for the 50*0.8*5
and the A(t) is 1,000
Hope i could help :D
Equilateral triangle are also equiangular which means all angles are 60 degrees. thus our needed altitude is the leg of a right triangle with other leg 4 and hypotenuse 8. 4² +a² = 8²
a² = 48
a=√48 = 4√3
Answer:
How to solve your problem
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7
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-7y^{2}-2y^{2}+y^{3}y-2y+5y^{3}-2y
−7y2−2y2+y3y−2y+5y3−2y
Simplify
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Combine exponents
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7
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5
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-7y^{2}-2y^{2}+{\color{#c92786}{y^{3}y}}-2y+5y^{3}-2y
−7y2−2y2+y3y−2y+5y3−2y
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7
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4
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-7y^{2}-2y^{2}+{\color{#c92786}{y^{4}}}-2y+5y^{3}-2y
−7y2−2y2+y4−2y+5y3−2y
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Combine like terms
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7
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4
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5
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{\color{#c92786}{-7y^{2}}}{\color{#c92786}{-2y^{2}}}+y^{4}-2y+5y^{3}-2y
−7y2−2y2+y4−2y+5y3−2y
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9
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4
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{\color{#c92786}{-9y^{2}}}+y^{4}-2y+5y^{3}-2y
−9y2+y4−2y+5y3−2y
3
Combine like terms
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9
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4
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-9y^{2}+y^{4}{\color{#c92786}{-2y}}+5y^{3}{\color{#c92786}{-2y}}
−9y2+y4−2y+5y3−2y
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9
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4
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-9y^{2}+y^{4}{\color{#c92786}{-4y}}+5y^{3}
−9y2+y4−4y+5y3
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Rearrange terms
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9
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4
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4
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{\color{#c92786}{-9y^{2}+y^{4}-4y+5y^{3}}}
−9y2+y4−4y+5y3
4
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9
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{\color{#c92786}{y^{4}+5y^{3}-9y^{2}-4y}}
y4+5y3−9y2−4y
Solution
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5
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9
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The perimeter of the polygon will be given by:
Perimeter=distance around the figure
Thus the perimeter will be:
P=9.9+(9.9-5.9)+5.9+15.9+(15.9-4.6)+4.6
P=9.9+4+5.9+15.9+11.3+4.6
P=51.6
Answer: 51.6 units
It is 8.525. what u do is 21÷40 and get .525 and then u add the 8 in the front and get 8.525
