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guapka [62]
3 years ago
9

A certain colony of bacteria triples in length every 10 minutes. It's length is now 1 millimeter. how long will it be in 40 minu

tes?
Mathematics
1 answer:
iren [92.7K]3 years ago
5 0
So we know that the bacteria grows 3x as much every ten minutes.  From this we know that it'll be 3 mm in 20 minute.  Continuing the pattern, we can figure out that in 40 minutes the bacteria will be 81 mm.

The pattern: 10min - 3  20min - 9  30min - 27  40min - 81
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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
Josh is hiking glacier national park. He has now hiked a total of 17km and is 2km short of being 1/2 of the way done with his hi
Alla [95]

Answer: The total length in kilometers is 38 km


Step-by-step explanation:


17 + 2 = 19 km is 1/2 of the total length, that is,


19 = (1/2)h


(1/2)h = 19 multiply both sides by 2


h = 2*19


h = 38 km

5 0
3 years ago
4w^3−8w+12<br> factor the polynomial<br><br> 5t^2+20t+50<br> factor the polynomial
DanielleElmas [232]
<span>4w^3−8w+12 = 4 (w^3 - 2w + 3)

</span><span>5t^2+20t+50 = 5 (t^2 + 4t + 10)</span>
7 0
4 years ago
Verify that TUV ~ XYZ. Find the scale factor of TUV to XYZ.
Sonbull [250]

Answer:

35

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The length of a concrete slab is three more than three times the width. It's area is 330 square feet. What is the length of the
marin [14]

Answer:

Width: 10.5 feet

Length: 31.5 feet

Step-by-step explanation:

Let x represent width of the concrete slab.

We have been given that the length of a concrete slab is three more than three times the width. So length of the slab would be 3x.

We are also told that the area of slab is 330 square feet. We can represent this information in an equation as:

x\cdot 3x=330

3x^2=330

x^2=\frac{330}{3}

x^2=110

Now, we will take square root of both sides.

\sqrt{x^2}=\sqrt{110}

x=10.488\approx 10.5

Therefore, the width of slab is approximately 10.5 feet.

The length of the slab would be 3x\Rightarrow3(10.5)=31.5.

Therefore, the length of slab is approximately 31.5 feet.

6 0
3 years ago
Read 2 more answers
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