Surface Area of a Cube = 6 a 2
(a is the length of the side of each edge of the cube)
In words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a 2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.
Surface Area of a Rectangular Prism = 2ab + 2bc + 2ac
(a, b, and c are the lengths of the 3 sides)
In words, the surface area of a rectangular prism is the area of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same.
The area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab.
Surface Area of Any Prism
(b is the shape of the ends)
Surface Area = Lateral area + Area of two ends
(Lateral area) = (perimeter of shape b) * L
Surface Area = (perimeter of shape b) * L+ 2*(Area of shape b)
Surface Area of a Sphere = 4 pi r 2
(r is radius of circle)
Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h
(h is the height of the cylinder, r is the radius of the top)
Surface Area = Areas of top and bottom +Area of the side
Surface Area = 2(Area of top) + (perimeter of top)* height
Surface Area = 2(pi r 2) + (2 pi r)* h
In words, the easiest way is to think of a can. The surface area is the areas of all the parts needed to cover the can. That's the top, the bottom, and the paper label that wraps around the middle.
You can find the area of the top (or the bottom). That's the formula for area of a circle (pi r2). Since there is both a top and a bottom, that gets multiplied by two.
The side is like the label of the can. If you peel it off and lay it flat it will be a rectangle. The area of a rectangle is the product of the two sides. One side is the height of the can, the other side is the perimeter of the circle, since the label wraps once around the can. So the area of the rectangle is (2 pi r)* h.
Add those two parts together and you have the formula for the surface area of a cylinder.
Surface Area = 2(pi r 2) + (2 pi r)* h
Answer:
7.5×10^4
Step-by-step explanation:
Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.
Hope this helps <3
Answer:
x = 5
Step-by-step explanation:
Using the fact that line segments AB and BC are parts of the whole line segment AC, we can write the following equation:
AB + BC = AC
Now, using the given values, we can substitute in for the equation and solve for x:
AB + BC = AC
9 + 2x - 5 = x + 9
2x + 4 = x + 9
x = 5
Thus, we have found that for these sets of equations for these line segments, our value for x should be 5.
Cheers.
Answer:
CD = 14 cm; DE = 21 cm
Step-by-step explanation:
The perimeter is the sum of side lengths (in centimeters), so ...
CD + DE + CF + EF = 55
CD + DE + 8 + 12 = 55 . . . . . . . substittute for CF and EF
CD + DE = 35 . . . . . . . . . . . . . . subtract 20
___
The segment DF is a diagonal of the rhombus, so bisects angle D. That angle bisector divides ΔCDE into segments that are proportional. That is, ...
CD/DE = CF/EF = 8/12 = 2/3
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So, we have two segments whose sum is 35 (cm) and whose ratio is 2 : 3. The total of "ratio units is 2+3=5, so each must stand for a length unit of 35/5 = 7 (cm). The sides are ...
CD = 2·7 cm = 14 cm
DE = 3·7 cm = 21 cm
<em>Check</em>
CD + DE = (14 +21) cm = 35 cm . . . . . matches requirements
<u>ANSWER</u>
<u>EXPLANATION</u>
We want to solve the trigonometric equation;
, where .
Taking square root of both sides gives,
Since B is in the first quadrant, we proceed with;
, because the tangent ratio is positive in the first quadrant.
.
Correct to the nearest degree,