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baherus [9]
3 years ago
9

A garden is a rectangle 100 feet long and 50 feet wide. The portion of the garden devoted to herbs is 10 feet by 10 feet. The re

st of the area is for vegetables. How many square feet are given for growing vegetables?
Mathematics
2 answers:
postnew [5]3 years ago
4 0

Answer:

Area for vegetable = 4900  square ft

Step-by-step explanation:

The garden is a rectangle . According to the question the length of the garden is 100 ft and the width is 50 ft.  The portion of area devoted to herb is 10 ft by 10 ft. That means the area devoted to herbs is a square. The area devoted for vegetable can be computed when one subtract area for herbs from the area of the garden.

Area of the rectangle garden = length × width

length = 100 ft

width = 50 ft

Area of the rectangle garden = 100 × 50

Area of the rectangle garden = 5000 square ft

Area devoted for herbs (square) = length²

length = 10 ft

Area devoted for herbs (square) = 10²

Area devoted for herbs (square) = 10 × 10

Area devoted for herbs (square) = 100 square ft

Area for vegetable = Area of garden - Area for herbs

Area for vegetable = 5000 - 100

Area for vegetable = 4900  square ft

lys-0071 [83]3 years ago
3 0
4,900 sq feet are given for growing vegetables
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6 0
3 years ago
The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m  + 40 m  + c = 120 m

61 m +  c = 120 m

c = 120 m - 61 m

c = 59 m

                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

8 0
3 years ago
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The distributive property: a(b + c) = ab + ac


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7 0
4 years ago
Read 2 more answers
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