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3 years ago

How do you find the solution

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

4 0

Given that there were 30 bacteria present originally.
a=30
Also given that the number doubles every one hour.
After one hour the count =2×30=60
Since the count doubles every hour it forms a G.P. with r=2
a,ar,ar2........
The count at the end of 2nd hour=ar^2=30×2^2=120
The count at the end of 4th hour=ar^4=30×2^4=480
The count at the end of 8th hour=ar^8n=30×2^8=7680
etc.
The count at the end of n^th hour=ar^n=30×2^n

Also, the equation y=30*(2)^8/1 give the same result y= 7680

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sergejj [24]

Answer: 38

Step-by-step explanation: to evaluate an expression for these variables, plug in those numbers for the variables.

Plugging in -1 for y and -19 for z gets you this:

-2(-1)^2(-19)

Now simplify according to order of operations (PEMDAS - parenthesis exponents multiplication/division addition/subtraction)

-1^2 = 1

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-2(-19) = 38

Your answer is 38

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2 years ago

Find the length of the following two-dimensional curve. r (t ) = (1/2 t^2, 1/3(2t+1)^3/2) for 0 < t < 16

andrezito [222]

Answer:

r = 144 units

Step-by-step explanation:

The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;

$r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)} dt$

In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.

Substituting the terms of the equation and the derivative of r´, as follows,

$r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)} dt$

Doing the operations inside of the brackets the derivatives are:

1 ) $(\frac{d((1/2)t^2)}{dt} )^2= t^2$

2) $\frac{(d(1/3)(2t+1)^{3/2})}{dt} )^2=2t+1$

Entering these values of the integral is

$r(t)= \int\limits^{16}_{0} \sqrt{t^2 +2t+1} dt$

It is possible to factorize the quadratic function and the integral can reduced as,

$r(t)= \int\limits^{16}_{0} (t+1) dt= \frac{t^2}{2} + t$

Thus, evaluate from 0 to 16

$\frac{16^2}{2} + 16$

The value is $r= 144 units$

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3 years ago

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mina [271]

Oh shucks, that's crazy man

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3 years ago

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3 years ago

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Makovka662 [10]

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2 years ago