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g100num [7]
3 years ago
14

How many times will we move the decimal in the dividend? 0.9) 6,815.3​

Mathematics
1 answer:
son4ous [18]3 years ago
6 0

Step-by-step explanation:

So when you divide by powers of ten (ten times itself some number of times, like ten, one hundred, one thousand, etc.), the decimal moves to the left. For each power of ten, you move one place to the left. So if you divided a number by one hundred (10^2 or 10*10), you would move two places to the left.

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Ratling [72]

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Then the side

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Which makes choice C the correct choice.

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What are the discontinuities of the function f(x) = the quantity x squared minus 16 over the quantity 4x minus 24?
ANEK [815]
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3 years ago
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mestny [16]
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6 0
3 years ago
PLEASE HELP A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwi
il63 [147K]
To solve our questions, we are going to use the kinematic equation for distance: d=vt
where
d is distance 
v is speed  
t is time 

1. Let v_{w} be the speed of the wind, t_{w} be time of the westward trip, and t_{e} the time of the eastward trip. We know from our problem that the distance between the cities is 2,400 miles, so d=2400. We also know that the speed of the plane is 450 mi/hr, so v=450. Now we can use our equation the relate the unknown quantities with the quantities that we know:

<span>Going westward:
The plane is flying against the wind, so we need to subtract the speed of the wind form the speed of the plane:
</span>d=vt
2400=(450-v_{w})t_{w}

Going eastward:
The plane is flying with the wind, so we need to add the speed of the wind to the speed of the plane:
d=vt
2400=(450+v_{w})t_{e}

We can conclude that you should complete the chart as follows:
Going westward -Distance: 2400 Rate:450-v_w Time:t_w
Going eastward -Distance: 2400 Rate:450+v_w Time:t_e

2. Notice that we already have to equations:
Going westward: 2400=(450-v_{w})t_{w} equation(1)
Going eastward: 2400=(450+v_{w})t_{e} equation (2)

Let t_{t} be the time of the round trip. We know from our problem that the round trip takes 11 hours, so t_{t}=11, but we also know that the time round trip is the time of the westward trip plus the time of the eastward trip, so t_{t}=t_w+t_e. Using this equation we can express t_w in terms of t_e:
t_{t}=t_w+t_e
11=t_w+t_e equation
t_w=11-t_e equation (3)
Now, we can replace equation (3) in equation (1) to create a system of equations with two unknowns: 
2400=(450-v_{w})t_{w}
2400=(450-v_{w})(11-t_e) 

We can conclude that the system of equations that represent the situation if the round trip takes 11 hours is:
2400=(450-v_{w})(11-t_e) equation (1)
2400=(450+v_{w})t_{e} equation (2)

3. Lets solve our system of equations to find the speed of the wind: 
2400=(450-v_{w})(11-t_e) equation (1)
2400=(450+v_{w})t_{e} equation (2)

Step 1. Solve for t_{e} in equation (2)
2400=(450+v_{w})t_{e}
t_{e}= \frac{2400}{450+v_{w}} equation (3)

Step 2. Replace equation (3) in equation (1) and solve for v_w:
2400=(450-v_{w})(11-t_e)
2400=(450-v_{w})(11-\frac{2400}{450+v_{w}} )
2400=(450-v_{w})( \frac{4950+11v_w-2400}{450+v_{w}} )
2400=(450-v_{w})( \frac{255011v_w}{450+v_{w}} )
2400= \frac{1147500+4950v_w-2550v_w-11(v_w)^2}{450+v_{w}}
2400(450+v_{w})=1147500+2400v_w-11(v_w)^2
1080000+2400v_w=1147500+2400v_w-11(v_w)^2
(11v_w)^2-67500=0
11(v_w)^2=67500
(v_w)^2= \frac{67500}{11}
v_w= \sqrt{\frac{67500}{11}}
v_w=78

We can conclude that the speed of the wind is 78 mi/hr.
6 0
3 years ago
What is 100.34567 in a fraction and word form? Thank you! ❤
amid [387]
Wait sorry in fraction for it is 1000000/10 
6 0
3 years ago
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