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jeka57 [31]
3 years ago
12

19/20 as terminating decimal

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0
19/20

= 15/20 + 4/20

= 3/4 + 1/5

= 0.75 + 0.20

= 0.95
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Given the position of the particle, what the position(s) of the particle when it’s at rest
choli [55]

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X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t

The velocity function is the derivative of the position:

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\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}

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\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

undefined

6 0
1 year ago
If 32^2c=8^c what is the value of c? 1 2 3 5
Elina [12.6K]

Answer:

  • as written, c ≈ 0.000979 or c = 4
  • alternate interpretation: c = 0

Step-by-step explanation:

<em>As written</em>, you have an equation that cannot be solved algebraically.

  (32^2)c = 8^c

  1024c = 8^c

  1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero

A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.

___

If you mean ...

  32^(2c) = 8^c

  (2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2

  2^(10c) = 2^(3c) . . . . . . . simplify

  10c = 3c . . . . . . . . . . . . . .log base 2

  7c = 0 . . . . . . . . . . . . . . . subtract 3c

  c = 0 . . . . . . . . . . . . . . . . divide by 7

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