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Mariulka [41]
3 years ago
10

Can anyone help me pls

Mathematics
1 answer:
Nikitich [7]3 years ago
3 0

9514 1404 393

Answer:

  A. some values (x = 4)

  B. all values

  C. no values

Step-by-step explanation:

If you subtract the right-side expression from both sides of an equation, so you get something of the form ...

  f(x) = 0

Then there are three possibilities:

  f(x) is some function of x . . . . "some values" make it true*

  f(x) = 0 . . . . "all values" make it true; the equation is an identity

  f(x) = constant (non-zero) . . . . "no values" of x will make it true

___

A: 3x = x +8

  3x -(x +8) = 0

  2x -8 = 0 . . . . . . . . true for "some" values of x, namely, x = 4

__

B: 5x -(x +x +x +x +x) = 0

  0 = 0 . . . . . . . . . . . true for "all" values of x

__

C: 2(x +5) -(2x +5) = 0

  2x + 10 -2x -5 = 0

  5 = 0 . . . . . . . . . . . true for "no values" of x (always FALSE)

_____

* The possibility exists that the function f(x) does not have 0 in its range. In that case, there are no real solutions for the equation. For example |x|+1 = 0 has no solutions. For expressions linear in x, there will always be a solution.

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kodGreya [7K]

Answer: 25.5 m^{3}

Step-by-step explanation:

If the volume of a sphere is

V=\frac{4}{3} \pi r^{3}

Where r=2.3 m is the radius

The volume of a hemisphere is half the volume of the total sphere:

V_{hemisphere}=\frac{V}{2}=\frac{\frac{4}{3} \pi r^{3}}{2}

V_{hemisphere}=\frac{2}{3} \pi r^{3}

Solving this equation:

V_{hemisphere}=\frac{2}{3} \pi (2.3 m)^{3}

Finally:

V_{hemisphere}=25.48m^{3} \approx 25.5 m^{3}

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Musya8 [376]

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9. disagree;

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