24ft long, 12ft wide
2/.25=8 8x3=24
1/.25=4 4x3=12
Answer:
P’(-1, 1), I’ (1, 2), G’(1,0)
Step-by-step explanation:
Answer:
See below for answers and explanations
Step-by-step explanation:
Top left: Since y can't be greater than 0 but is equal to 0, then the range is (-∞,0] and the domain is (-∞,∞) since there are no domain restrictions
Top right: Since both x and y have no restrictions, then the domain is (-∞,∞) and the range is (-∞,∞)
Bottom left: Since y cannot be less than 2 but equal to it, and x holds no domain restrictions, then the domain is (-∞,∞) and the range is [2,∞)
Bottom right: Since both x and y have no restrictions, then the domain is (-∞,∞) and the range is (-∞,∞)
Answer:
3. 150.72 in²
4. 535.2cm²
Step-by-step Explanation:
3. The solid formed by the net given in problem 3 is the net of a cylinder.
The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.
The surface area = Area of the 2 circles + area of the rectangle
Take π as 3.14
radius of circle = ½ of 4 = 2 in
Area of the 2 circles = 2(πr²) = 2*3.14*2²
Area of the 2 circles = 25.12 in²
Area of the rectangle = L*W
width is given as 10 in.
Length (L) = the circumference or perimeter of the circle = πd = 3.14*4 = 12.56 in
Area of rectangle = L*W = 12.56*10 = 125.6 in²
Surface area of net = Area of the 2 circles + area of the rectangle
= 25.12 + 125.6 = 150.72 in²
4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)
= 
Where,
b = 8 cm
h = ![\sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A = 2(0.5*8*6.9) + 3(20*8)](https://tex.z-dn.net/?f=%20%5Csqrt%7B8%5E2%20-%204%5E2%7D%20%3D%20%5Csqrt%7B48%7D%20%3D%206.9%20cm%7D%20%28Pythagorean%20theorem%29%3C%2Fp%3E%3Cp%3Ew%20%3D%208%20cm%3C%2Fp%3E%3Cp%3E%5Btex%5DS.A%20%3D%20%202%280.5%2A8%2A6.9%29%20%2B%203%2820%2A8%29)


