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Lorico [155]
3 years ago
15

Directions: Drag each operation and expression to the correct location on the equation. Not all operations or expressions will b

e used. One milliliter of blood contains approximately 5 million red blood cells, and one pint is about 473 milliliters. Create and solve an equation that can be used to find approximately how many red blood cells are in a pint of blood. Use scientific notation for the first box of the equation. × 2.4 × 109 9.73 × 106 2.4 × 106 5 × 106 473 4.73 5 × 109 + ÷ – 9.73 × 109
Mathematics
2 answers:
butalik [34]3 years ago
6 0
5.0 x 10^6 X 473 = 2.4 x 10^9

There are 5 x 10^6 blood cells in a ml, and 473 ml in a pint. simply multiply both numbers together and put the answer in scientific notation (one digit before the decimal)
bearhunter [10]3 years ago
3 0

Answer : The correct answer is, 2.63\times 10^{11}

Step-by-step explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.  The numerical digit lies between 0.1.... to 9.9.....

For example :

5000 is written as 5.0\times 10^3

889.9 is written as 8.899\times 10^{-2}

In this examples, 5000 and 889.9 are written in the standard notation and 5.0\times 10^3  and 8.899\times 10^{-2}  are written in the scientific notation.

As we are given the following conversions:

1mL=5\times 10^6RBC

1\text{ Pint}=437mL

Now we have to calculate the number of RBC present in a pint of blood.

As, 1mL=5\times 10^6RBC

and, 1\text{ Pint}=437mL

So, 437mL=\frac{437mL}{1mL}\times 5\times 10^6RBC=2365\times 10^6RBC=2.4\times 10^9RBC

Hence, the correct answer is, 2.4\times 10^9RBC

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Y = k/x
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sea level =0 A submarine is 120ft below sea level it climbs 50 ft towards sea level where the submarine comparrf to sea level an
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<h2>70 ft</h2>

Step-by-step explanation:

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9

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2 years ago
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A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
VMariaS [17]

Answer:

Q = arctan(7.1739) = 82.06

Step-by-step explanation:

Given:

- The mass of the person m = 20.3 kg

- The distance traveled up the ladder s = 1.1 m

- The gravitational constant g = 9.8 m/s^2

- The coefficient of static friction u_s = 0.23

- Total length of the ladder

Find:

The minimum angle θ, that would allow the person to climb without ladder slipping

Solution:

- Taking moments about point of ladder and wall contact A to be zero:

                   -F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0

- Taking Sum of vertical forces to be zero:

                    F_n,b - m*g = 0

                    F_n,b = m*g

- The frictional force F_f is given by:

                   F_f = u_s*F_n,b = u_s*m*g

- Plug the values back in:

                  - m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0

Simplify:

                  4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

                        6.6*cos(Q) = 4*u_s*sin(Q)

                               tan(Q) = 6.6 / 4*u_s

- Plug in the values:

                               tan(Q) = 6.6 / 4*0.23

                                    Q = arctan(7.1739) = 82.06

                   

                     

4 0
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