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jonny [76]
3 years ago
6

207,567 to the nearest hundred thousand

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
6 0
The hundred thousands place is the 2. Since the place below that is 0, it rounds down.
200,000
Kitty [74]3 years ago
6 0
207, 567 to the nearest hundred thousand is: 200, 000
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Solve in 5 + in 2x = 3
faltersainse [42]

\ln5+\ln 2x=3\\\\\ln10x=3\\\\e^3=10x\\\\x=\dfrac{e^3}{10}

3 0
2 years ago
You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40.
koban [17]

You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40. How much did each item cost?

 

First let’s define a shirt as X and pants as Y!

 

In the first scenario, the equation will be 3X + Y =$32

In the second scenario, the equation will be 2X + 3Y = $40  

 

Now because we have two different variables (X & Y) let’s solve for one of them (Ex: Y) to simplify the equations!

 

3X + Y= $32—> subtract 3X on both sides of the equation  

-3X.        - 3X

 

Net result: Y= 32- 3X

 

Now we can insert this into the second equation 2X + 3Y = $40 in order to solve for X variable  

 

2X + 3(32-3X) = $40

2X + 96 - 9X = $40 —> simplify the equation  

-7X + 96 = $40 —> substrate 96 from both sides  

-7X = -56 —> divide both sides by -7  

X= 8  

 

Now reinsert 8= X into 3X + Y = $32

3(8) + Y = $32

24+ Y = $32 —> substract 24 from both sides

y = 8

   

ANSWER: X= 8, Y= 8

If I am wrong I am sorry >_<

5 0
2 years ago
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
3 years ago
Read 2 more answers
Becky decorated 144 cakes in 6 days. how many she decorate in 1 day?
mario62 [17]

Answer: HOPE THIS HELPED! :D    ;P

24 cakes per day

Step-by-step explanation:

144 / 6 = 24

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<span>226.667%

tell me if its right...

</span>
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