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Marizza181 [45]
2 years ago
13

Find the surface area of the cylinder

Mathematics
1 answer:
STatiana [176]2 years ago
6 0
<h2><u>Solution</u>:-</h2>

• Surface area of a cylinder = 2πr(r + h) sq. units.

In the above diagram,

Radius (r) of the cylinder = 13 cm.

Height (h) of the cylinder = 39 cm.

<h3>• Taking value of π = 3.14</h3>

Hence, Surface area of the cylinder = 2 × 3.14 × 13(13 + 39)

Surface area of the cylinder = (81.64 × 52) cm²

Surface area of the cylinder = 4245.28 cm²

The surface area of the cylinder is <u>4</u><u>2</u><u>4</u><u>5</u><u>.</u><u>2</u><u>8</u><u> </u><u>cm²</u>. [Answer]

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If the diameter of the circle is above 20mm what is the area
Phoenix [80]

Answer:

The area would be 314.16 mm squared

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3 years ago
Solving multi step equations
hichkok12 [17]

The no-brain way to do it is to

  1. subtract one side from both sides so you have <em>(something) = 0</em>.
  2. divide by the coefficient of the variable.
  3. add the opposite of the constant.

Of course, at some point, you need to simplify the equation so you have something like

... ax + b = 0 . . . . . . where <em>a</em> and <em>b</em> are some constants that may be positive or negative

17) Subtract the right side.

... 10(x +3) -(-9x -4) -(x -5 +3) = 0

... 10x + 30 +9x +4 -x +5 -3 = 0 . . . . . eliminate parentheses

... x(10+9-1) +(30+4+5-3) = 0 . . . . . . . collect terms

... 18x +36 = 0 . . . . . . . . . . . . . . . . . . . simplified

... x + 2 = 0 . . . . . . . . . . . . . . . . . . . . . .divide by 18, the coefficient of x

... x = -2 . . . . . . . . . . . . . . . . . . . . . . . . add the opposite of the constant

19) Add the opposite of the left side.

... 0 = -9(1 +7x) +12(x -12)

... 0 = -9 -63x +12x -144 . . . . . . eliminate parenthses

... 0 = -51x -153 . . . . . . . . . . . . . simplify

... 0 = x +3 . . . . . . . . . . . . . . . . . divide by -51, the coefficient of x

... -3 = x . . . . . . . . . . . . . . . . . . . add the opposite of the constant

_____

If you examine the variable's coefficients you can make a choice of side to subtract that results in a positive coefficient of the variable.

This method puts variable and constant together until the end. The approach usually taught is to separate the variable terms and constant terms. (The number of steps required is the same either way.)

The reason this is "no brain" is that it always works and requires no judgment as to what you add or subtract from where. Applying a little judgment as described above can make it so you're mostly working with positive numbers, but the method works whether the numbers are positive or negative.

6 0
3 years ago
Angle 1 and Angle 2 are complementary angles. Angle 1 is 5x degrees. Angle 2 is 50 degrees. Find the value of x. Type ONLY the n
jenyasd209 [6]

Answer:

x = 8

Step-by-step explanation:

complementary angles sum to 90°

sum the 2 angles and equate to 90

5x + 50 = 90 ( subtract 50 from both sides )

5x = 40 ( divide both sides by 5 )

x = 8

3 0
1 year ago
Help me with this problem
ryzh [129]
The answer is 60 since the other shape has same answer for that side
8 0
2 years ago
Read 2 more answers
A radio telescope has a parabolic surface, as shown below.
krek1111 [17]
OK, so the graph is a parabola, with points x=0,y=0; x=6,y=-9; and x=12,y=0

Because the roots of the equation are 0 and 12, we know the formula is therefore of the form

y = ax(x - 12), for some a

So put in x = 6

-9 = 6a(-6)

9 = 36a

a = 1/4

So the parabola has a curve y = x(x-12) / 4, which can also be written y = 0.25x² - 3x

The gradient of this is dy/dx = 0.5x - 3

The key property of a parabolic dish is that it focuses radio waves travelling parallel to the y axis to a single point. So we should arrive at the same focal point no matter what point we chose to look at. So we can pick any point we like - e.g. the point x = 4, y = -8

Gradient of the parabolic mirror at x = 4 is -1

So the gradient of the normal to the mirror at x = 4 is therefore 1.

Radio waves initially travelling vertically downwards are reflected about the normal - which has a gradient of 1, so they're reflected so that they are travelling horizontally. So they arrive parallel to the y axis, and leave parallel to the x axis.

So the focal point is at y = -8, i.e. 1 metre above the back of the dish.
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