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)



<h3>
Answer: 9</h3>
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Work Shown:
Apply the pythagorean theorem
a^2 + b^2 = c^2
12^2 + b^2 = 15^2
144 + b^2 = 225
b^2 = 225-144
b^2 = 81
b = sqrt(81)
b = 9
Answer:
6:1:2
Step-by-step explanation:
Let a = Able's score, b = Ben's score, and c = Cal's score.
Since
Able's score was 6 times Ben's score, that means a = 6b.
Cal's score was a third of Able's score, so that means c = a/3. And since a = 6b, that means c = 6b / 3 = 2b.
Thus, the ratio of Able's score to Ben's score to Cal's score, a:b:c, is 6:1:2, because c is twice as much as b and a is 6 times as much as b.
You would have to make a point at 6/3 and go down one and to the left 7... then your answer would be -1/2 sooo D=-1 hope this helps