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NikAS [45]
3 years ago
14

Subtract5x2 + 10x 1 from 6x2 x + 3.

Mathematics
1 answer:
mixer [17]3 years ago
3 0

Answer:

-6x^{3} +5x^{2} +10x+3

Step-by-step explanation:

5x^2+10x^1−6x^2x+3

=5x^2+10x+−6x^3+3

Answer:

=−6x^3+5x^2+10x+3

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Solve1 over 25 = 5^x ^+ 4. <br> x = -7/2<br> x = −6 <br> x = 9/2 <br> x = 2
Basile [38]
The equation seems to be

   1
-------- = 5 ^ (x + 4)
  25

In that case, this is the solution, stept by step:

1) factor 25: 25 = 5^2

          1
=> ----------- = 5^(x + 4)
         5^2

2) invert 5^2

=> 5^(-2) = 5^(x+4)

3) Given that the bases are equal, the exponents also have to be equal:

=> - 2 = x + 4

3) transpose +4:

=> - 2 - 4 = x

=> x = - 6

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6 0
3 years ago
Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3. What is the volume of the
Arisa [49]
:<span>  </span><span>You need to know the derivative of the sqrt function. Remember that sqrt(x) = x^(1/2), and that (d x^a)/(dx) = a x^(a-1). So (d sqrt(x))/(dx) = (d x^(1/2))/(dx) = (1/2) x^((1/2)-1) = (1/2) x^(-1/2) = 1/(2 x^(1/2)) = 1/(2 sqrt(x)). 

There is a subtle shift in meaning in the use of t. If you say "after t seconds", t is a dimensionless quantity, such as 169. Also in the formula V = 4 sqrt(t) cm3, t is apparently dimensionless. But if you say "t = 169 seconds", t has dimension time, measured in the unit of seconds, and also expressing speed of change of V as (dV)/(dt) presupposes that t has dimension time. But you can't mix formulas in which t is dimensionless with formulas in which t is dimensioned. 

Below I treat t as being dimensionless. So where t is supposed to stand for time I write "t seconds" instead of just "t".

Then (dV)/(d(t seconds)) = (d 4 sqrt(t))/(dt) cm3/s = 4 (d sqrt(t))/(dt) cm3/s = 4 / (2 sqrt(t)) cm3/s = 2 / (sqrt(t)) cm3/s. 

Plugging in t = 169 gives 2/13 cm3/s.</span>
4 0
3 years ago
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