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2 years ago

What is the volume of a ball with a diameter of 7 centimeters?

Mathematics

1 answer:

Leviafan [203]2 years ago

7 0

42.88. You might want to wait for another answer jyst to be 100% sure:)

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There is a lock with 5 buttons numbered from 1 to 5. The combination that opens this lock can be set to any sequence of buttons

Fittoniya [83]

Answer:

Answer Counterclockwise rotation about the origin by 180 degrees followed by reflection about the y-axis explanation the given figure has vertices p(1,2),q(2,1),r(3,2),s(3,3)recall that for counterclockwise rotation about the origin, (x,y)\rightarrow (-x,-y)when we rotate figure pqrs counterclockwise through an angle of 180° about the origin, the coordinates will become p1( - 1, - 2),q1( - 2, - 1),r1( - 3, - 2),s1( - 3, - 3)next, we reflect in the y-axis by negating the x-coordinates of the resulting figure to obtain,p'(1, - 2),q'(2, - 1),r'(3, - 2),s'(3, - 3)

Step-by-step explanation:

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3 years ago

Which expression is equivalent to

cestrela7 [59]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$