PLEASE HELP MATH MATH MATH PLEASE HELP HELP PLEASEE
Recall: We use perimeter when building a fence around a yard or any place that needs to be enclosed. in that case we would measure the distance in feet, yards, or meters. In Math, we usually measure the perimeter of polygons. To find the perimeter of any polygon, we add the lengths of the sides.
Show your solution on the space provided in each item. arrange the terms of your final answer in STANDARD FORM.
1.)Solve for a simplified polynomial expression that represents the total length of the south side of the field.
2.) Write and simplify a polynomial expression that represents the perimeter of the eggplant field. State one reason why the perimeter would be useful to Mang Tony.
3.) solve for the area of mang tani squash field if x=3 and y =7. (A=lw)
4.) mang tani would like he's been plants to grow to a height of (x+3)feet.write and simplify polynomial expression that represents the volume of the beam plants if they reach a height of (x+3) feet. (V=lwh)
5.) Mang Tony realized he forgot to include an okra field into his field layout. He plans to use half the length and half the width of the pitch if you'd another to plant okra. Solve for a polynomial expression that represents the area of the new okra field. (A=lw)
Recall: We use perimeter when building a fence around a yard or any place that needs to be enclosed. in that case we would measure the distance in feet, yards, or meters. In Math, we usually measure the perimeter of polygons. To find the perimeter of any polygon, we add the lengths of the sides.
Show your solution on the space provided in each item. arrange the terms of your final answer in STANDARD FORM.
1.)Solve for a simplified polynomial expression that represents the total length of the south side of the field.
2.) Write and simplify a polynomial expression that represents the perimeter of the eggplant field. State one reason why the perimeter would be useful to Mang Tony.
3.) solve for the area of mang tani squash field if x=3 and y =7. (A=lw)
4.) mang tani would like he's been plants to grow to a height of (x+3)feet.write and simplify polynomial expression that represents the volume of the beam plants if they reach a height of (x+3) feet. (V=lwh)
5.) Mang Tony realized he forgot to include an okra field into his field layout. He plans to use half the length and half the width of the pitch if you'd another to plant okra. Solve for a polynomial expression that represents the area of the new okra field. (A=lw)
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Problem:
1.) Solve for a simplified polynomial expression that represents the total length of the south side of the field.
Solution:
Length of the South field
L = (5x + 2) + (x² - 9) + (x² - 7x + 12)
L = x² + x² + 5x - 7x + 2 - 9 + 12
L = 2x² - 2x + 5 (feet)
2.) Write and simplify a polynomial expression that represents the perimeter of the eggplant field. State one reason why the perimeter would be useful to Mang Tony.
Solution:
Perimeter of eggplant field
P = (5x - 2) + (5x - 2) + (5x + 2) + (5x + 2)
P = 20x (feet)
Why the perimeter would be useful to Mang Tony.
The perimeter would be useful to Mang Tony because he could easily enclosed and layout how much area he required for every plant he wished to place in the field
3.) Solve for the area of Mang Tony squash field if x=3 and y =7. (A=lw)
Solution:
Squash field Area
if x = 3 ; if y = 7
A = (xy - 1)(xy + 1)
A = ((3)(7) - 1))((3)(7) + 1))
A = (21 - 1)(21 + 1)
A = 20(22)
A = 440 feet²
4.) Mang Tony would like he's been plants to grow to a height of (x+3)feet. Write and simplify polynomial expression that represents the volume of the beans plants if they reach a height of (x+3) feet. (V=lwh)
Solution:
Volume of the beans plants
L = (5x + 2) ; W = 4x ; H = (x + 3)
V = (5x + 2)(4x)(x + 3)
V = 4x(5x² + 17x + 6)
V = 20x³ + 68x² + 24x (feet³)
5.) Mang Tony realized he forgot to include an okra field into his field layout. He plans to use half the length and half the width of the petchay if you'd another to plant okra. Solve for a polynomial expression that represents the area of the new okra field. (A=lw)
Solution:
Petchay field L = 6x ; W = 4x
Area of petchay field = 6x(4x)
Area of petchay field = 24x²
Half Area of petchay field = 24x²/2
Half Area of petchay field = 12x² (feet²)
Area of the new okra field is half of the petchay field
Area of the new okra field = 24x²/2
Area of the new okra field = 12x² (feet²)
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