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4 years ago

Identify five common business uses for electronic spreadsheets

2 answers:

4 0

For the answer to the question above
Forecasting how a business might do in the future.
Calculating tax.
Doing basic payrolls.
Calculating Revenues.
Producing charts.

--Going past 5--

Inventory tracking
Very (VERY) basic CRM for small businesses
I hope my answer helped you.

tiny-mole [99]4 years ago

3 0

Electronic spreadsheet uses vary from simple calculations to complex numerical solutions. In business, it may be used for (1) organizing statistical and financial information, (2) income statements, (3) balance sheets, (4) frequency distribution tables, (5) creative presentation.

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A cyclist, while overtaking another bike, increases his speed uniformly from 4.2 mis to 6.3 m/s over a time interval of 5.3 s. H

avanturin [10]

Answer:

27.82m

Explanation:

The distance of the cyclist can be calculated using the formula:

S = ut + 1/2 (at^2)

Where; S = distance

u = initial velocity

a = acceleration

t = time

However, to find acceleration, we use:

a = v - u/t, where v is the final velocity

According to this question, u = 4.2m/s, v = 6.3m/s, t = 5.3s

a = 6.3 - 4.2 / 5.3

a = 2.1/5.3

a = 0.396m/s^2

Hence, using S = ut + 1/2 (at^2)

S = (4.2 × 5.3) + 1/2 (0.396 × 5.3 × 5.3)

S = 22.26 + 5.56

S = 27.82

Therefore, the cyclist travel in 27.82m

8 0

3 years ago

TRUE OR FALSE

stiks02 [169]

Answer:

I best answer is true sorry if it's wrong

3 0

3 years ago

Read 2 more answers

The average speed of a nitrogen molecule in air is about 6.70 ✕ 102 m/s, and its mass is about 4.68 ✕ 10-26 kg.(a) If it takes 3

Sveta_85 [38]

Answer:

a = 4.32*10^15 m/s^2

Explanation:

In order to calculate the acceleration of the nitrogen molecule, you take into account that the force experienced by the molecule, is equal to the change in its momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the nitrogen molecule

Δv: change in the speed of the molecule

Δt: time lapse of the change of the momentum of the molecule = 3.10*10^-13 s

Furthermore, by the Newton second law, you have:

$F=ma=m\frac{\Delta v}{\Delta t}\\\\a=\frac{\Delta v}{\Delta t}$ (2)

The change in the speed is given by:

$\Delta v=v-v_o=6.70*10^2m/s-(-6.70*10^2m/s)=1.34*10^3\frac{m}{s}$

where you have taken into account the direction of the initial and final speed.

You replace the values of all parameters in the equation (1) in order to calculate the acceleration:

$a=\frac{1.34*10^3m/s}{3.10*10^{-13}s}=4.32*10^{15}\frac{m}{s^2}$

The acceleration of the nitrogen molecule is 4.32*10^15m/s^2

5 0

3 years ago

HELP PLEASE

denis-greek [22]

B) Distance: the farther, the weaker!

4 0

4 years ago

Read 2 more answers

Emily takes a trip, driving with a constant velocity of 87.5 km/h to the north except for a 22 min rest stop. If Emily's average

rjkz [21]

<h2>Answer:</h2>

2.63 hours

<h2>Explanation:</h2>

First we have to read carefully the problem, It start telling us that Emily was driving with average velocity of 87.5 km/h to the north. Velocity is a physical variable that is definded in terms of :

Speed of the object (87.5 km/h)

Direction of the object (north)

Them the problem says that Emily's trip has an average velocity of 76.8 km/h to the north because she made a rest stop for 22 minutes.

Now you can ask yourself: If Emily was driving at 87.5 km/h, why the average speed of the trip was 76.8 km/h?

we need to see how speed is defined:

$Speed=\frac{distance}{time} =\frac{x}{y}$

***Note 1:

Distance will be called as "x" and units will be kilometers(km)

Time will be called as "y" and units will be hours(h)

So, 87.5 km/h would be the average speed for a distance "x" and a time "y" without stops. But Emily made a stop that took 22 minutes. For the same distance Emily's trip took more time and time is in the denominator. If our numerator is constant and our denominator gets higher, the final result will be lower

Now, we have the follow expressions:

$87.5=\frac{x}{y}~~(1) \\76.8=\frac{x}{y+\frac{22}{60}}~~(2)$

***Note 2:

we need to convert 22 minutes to hours. 1 hour= 60 minutes, so we need to apply the follow covert factor:

$y(hours)=y(minutes)*\frac{1(hour)}{60(minutes)}$

The problem aks us about how long does the trip take, it means that we have to find y.

We have two variables (x and y), and we have two equationts. We know that x have the same value for both problems (Because both average speeds have the same distance), so we can solve both equations for x and made equal each other

$87.5*y=x~(3)\\76.8*(y+\frac{22}{60})=x~(4)$

$87.5*y=76.8*(y+\frac{22}{60})~(5)$

We have to expand (5) and then we have to solve for y

$87.5*y=76.8*y+76.8*\frac{22}{60}~~(6)\\87.5*y-76.8*y=76.8*\frac{22}{60}~~(7)\\10.7*y=76.8*\frac{22}{60}~~(8)\\y=\frac{76.8}{10.7}*\frac{22}{60}(hours)~~(9)\\y=2.63(hours)~~(10)$

3 0

3 years ago