Solve using elimination. -2x + y = 1 -6x - y = -9 HELP
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Hello skm413002!
Apply the distributive property:
-3(x+2) =
By applying the distributive property, we can solve it as....
- Distributive property.
- Multiply -3 with x & 2.
- Add the 2 products.
- You'll get the answer as <u>-3x </u><u>-</u><u> </u><u>6</u><u>.</u>
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Hope it'll help you!
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Answer:
1. The amount of ice needed = 18 m²
2. The amount of fabric needed to manufacture the umbrella is 0.76 m²
3. The height of the cone, is 3.75 cm
4. The dimensions of the deck are;
Width = 28/3 m, breadth = 28/3 m
The area be 87.11 m²
5. The dimensions of the optimal design for setting the storage area at the corner, we have;
Width = 10m
Breadth = 10 m
The dimensions of the optimal design for setting the storage area at the back of their building are;
Width = 7·√2 m
Breadth = 7·√2 m
Step-by-step explanation:
1. The amount of ice needed is given by the volume, V, of the pyramid given by V = 1/3 × Base area × Height
The base area = Base width × Base breadth = 3 × 5 = 15 m²
The pyramid height = 3.6 m
The volume of the pyramid = 1/3*15*3.6 = 18 m²
The amount of ice needed = 18 m²
2. The surface area of the umbrella = The surface area of a cone (without the base)
The surface area of a cone (without the base) = π×r×l
Where:
r = The radius of the cone = 0.4 m
l = The slant height = √(h² + r²)
h = The height of the cone = 0.45 m
l = √(0.45² + 0.4²) = 0.6021 m
The surface area = π×0.4×0.6021 = 0.76 m²
The surface area of a cone (without the base) = 0.76 m²
The surface area of the umbrella = 0.76 m²
The amount of fabric needed to manufacture the umbrella = The surface area of the umbrella = 0.76 m²
3. The volume, V, of the cone = 1/3×Base area, A, ×Height, h
The volume of the cone V = 150 cm³
The base area of the cone A = 120 cm²
Therefore we have;
V = 1/3×A×h
The height of the cone, h = 3×V/A = 3*150/120 = 3.75 cm
4. Given that the deck will have railings on three sides, we have;
Maximum dimension = The dimension of a square as it is the product of two equal maximum obtainable numbers
Therefore, since the deck will have only three sides, we have that the length of each side are equal and the fourth side can accommodate any dimension of the other sides giving the maximum dimension of each side as 28/3
The dimensions of the deck are width = 28/3 m, breadth = 28/3 m
The area will then be 28/3×28/3 = 784/9 = =87.11 m²
5. The optimal design for setting the storage area at the corner of their property with four sides is having the dimensions to be that of of a square with equal sides of 10 m each as follows;
Width = 10m
Breadth = 10 m
The optimal design to have the storage area at the back of their building having a fence on only three sides, is given as follows;
Storage area specified = 98 m²
For optimal use of fencing, we have optimal side size of fencing = s = Side length of a square
s² = 98 m²
Therefore, s = √98 = 7·√2 m
Which gives the width = 7·√2 m and the breadth = 7·√2 m.