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ohaa [14]
3 years ago
9

The average human body cell is about 5.1x10^-4 cm in diameter. The diameter of a plant cell is 8x10^-3. What is the combine diam

eter of the body cell and plant cell?
Mathematics
1 answer:
suter [353]3 years ago
8 0

The combine diameter of the body cell and plant cell is 8.51 \times 10^{-3}

<u>Solution:</u>

Given, The average human body cell is about 5.1 \times 10^{-4} cm in diameter.  

The diameter of a plant cell is 8 \times 10^{-3}

We have to find what is the combine diameter of the body cell and plant cell?

In order to add or subtract numbers in scientific notation, the powers of ten must have the same exponent.  

In this case the diameter of an average human body cell is a power of -4 and the diameter of a plant cell is a power of -3.  

In order to add them, we can move the decimal over for the body cell diameter:

5.1 \times 10^{-4}=0.51 \times 10^{-3}

\text { Now add both the values }=8 \times 10^{-3}+0.51 \times 10^{-3}=(8+0.51) \times 10^{-3}=8.51 \times 10^{-3}

Hence, the combined diameter is 8.51 \times 10^{-3}

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Answer:

The amount paid by each house is $11.25

Step-by-step explanation:

Let the amount paid by each house is $K

Number of neighbors you worked for on Friday = 4

So the amount paid by all of them = 4K

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similarly amount paid = 5K

Number of neighbors you worked for on Sunday = 3,

here amount paid = 3K

Also, in total the amount paid = $135

So, Amount paid on {Friday + Saturday +Sunday} = Total Amount

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6 0
3 years ago
A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a well that is 90 ft deep. The bucket is f
Lisa [10]

Answer:

a) Lim(0-inf)  Work = (36 - 0.1*xi )*dx

b) Work = integral( (36 - 0.1*xi ) ).dx

c) Work = 2835 lb-ft

Step-by-step explanation:

Given:

- The weight of the bucket W = 3 lb

- The depth of the well d = 90 ft

- Rate of pull = 2.5 ft/s

- water flow out at a rate of = 0.25 lb/s

Find:

A. Show how to approximate the required work by a Riemann sum (let x be the height in feet above the bottom of the well. Enter xi∗as xi)

B. Express the Integral

C. Evaluate the integral

Solution:

A.

- At time t the bucket is xi = 2.5*t ft above its original depth of 90 ft but now it hold only (36 - 0.25*t) lb of water at an instantaneous time t.

- In terms of distance the bucket holds:

                           ( 36 - 0.25*(xi/2.5)) = (36 - 0.1*xi )

- Moving this constant amount of water through distance dx, we have:

                            Work = (36 - 0.1*xi )*dx

B.

The integral for the work done is:

                           Work = integral( (36 - 0.1*xi ) ).dx

Where the limits are 0 < x < 90.

C.

- Evaluate the integral as follows:

                           Work = (36xi - 0.05*xi^2 )

- Evaluate limits:

                           Work = (36*90 - 0.05*90^2 )  

                            Work = 2835 lb-ft

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