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dybincka [34]
4 years ago
6

What the volume of a hemisphere with a radius of 10 cm.

Mathematics
1 answer:
Setler79 [48]4 years ago
7 0
The volume of a hemisphere is 2/3 pi R^3. 
     1) 2/3 x 3.14 x 10^3
     2)2/3 x 3.14 x 1000
     3)2/3 x 3140
     4) 2093.3
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Analyze the diagram below and answer the question that follows.
Nat2105 [25]

Answer:

C

Step-by-step explanation:

the cos of an angle of 45° is equal to √2/2~=0,7071

Do you want the full explanation?

4 0
3 years ago
If AB is 12, what is the length of A'B'?
lukranit [14]

Triangles ABC and A'B'C are similar. This means that the corresponding sides are in the same ratio, in this case:

\frac{BC}{B^{\prime}C}=\frac{AC}{A^{\prime}C}=\frac{AB}{A^{\prime}B^{\prime}}

Replacing with data:

\begin{gathered} \frac{9}{6}=\frac{12}{A^{\prime}B^{\prime}} \\ 9\cdot A^{\prime}B^{\prime}=12\cdot6 \\ A^{\prime}B^{\prime}=\frac{72}{9} \\ A^{\prime}B^{\prime}=8 \end{gathered}

7 0
1 year ago
Whats the answer??? Pleaseeee!!!!
kvasek [131]

To find the 20th term in this sequence, we can simply keep on adding the common difference all the way until we get up to the 20th term.

The common difference is the number that we are adding or subtracting to reach the next term in the sequence.

Notice that the difference between 15 and 12 is 3.

In other words, 12 + 3 = 15.

That 3 that we are adding is our common difference.

So we know that our first term is 12.

Now we can continue the sequence.

12 ⇒ <em>1st term</em>

15 ⇒ <em>2nd term</em>

18 ⇒ <em>3rd term</em>

21 ⇒ <em>4th term</em>

24 ⇒ <em>5th term</em>

27 ⇒ <em>6th term</em>

30 ⇒ <em>7th term</em>

33 ⇒ <em>8th term</em>

36 ⇒ <em>9th term</em>

39 ⇒ <em>10th term</em>

42 ⇒ <em>11th term</em>

45 ⇒ <em>12th term</em>

48 ⇒ <em>13th term</em>

51 ⇒ <em>14th term</em>

54 ⇒ <em>15th term</em>

57 ⇒ <em>16th term</em>

60 ⇒ <em>17th term</em>

63 ⇒ <em>18th term</em>

66 ⇒ <em>19th term</em>

<u>69 ⇒ </u><u><em>20th term</em></u>

<u><em></em></u>

This means that the 20th term of this arithemtic sequence is 69.

5 0
4 years ago
A square with sides of 3 squareroot 2 is inscribed in a circle. what is the area of one of the sectors formed by the radii to th
mrs_skeptik [129]
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2).  Draw this to verify this statement.  Note that the height of each such triangular area is (3 sqrt(2))/2.

So now we have the base and height of one of the triangular sections.

The area of a triangle is A = (1/2) (base) (height).  Subst. the values discussed above,   A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2).  Show that this boils down to A = 9/2.

You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section.   Doing the problem this way, we get (1/4) (3 sqrt(2) )^2.  Thus, 
A = (1/4) (9 * 2) = (9/2).  Same answer as before. 
4 0
4 years ago
Complete the square and write in function form f(x)=x^2+5x+3
Over [174]
F(x) = x^2 +5x + 3

     = (x + 2.5)^2 - 6.25 + 3

     = (x + 2.5)^2 - 3.25

That is vertex form. I am not sure what you mean by function form
6 0
3 years ago
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