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Ivan
2 years ago
14

Find the value of x in theproportion 6 :(x+2)=2:5​

Mathematics
1 answer:
dexar [7]2 years ago
7 0

Answer:

x=13

Step-by-step explanation:

6:(x+2)=2:5

6/x+2=2/5

cross multiply

2(x+2)=6×5

open the bracket

2x+4=30

2x=30-4

2x=26

divide by 2

2x/2=26/2

x=13

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3 years ago
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Nicole is playing a video game where each round lasts \dfrac{7}{12} 12 7 ​ start fraction, 7, divided by, 12, end fraction of an
Harrizon [31]

Total number of hours she schduled to play the game = 3\frac{3}{4} hours.

Let us convert mixed fraction into improper fraction

3\frac{3}{4} = \frac{3*3+4}{4} = \frac{13}{4} \ hours.

Duration of each round = \frac{7}{12}.

In order to find the number of rounds Nicole can play, we need to divide total number of hours by duration of each round.

\frac{13}{4} ÷ \frac{7}{12}

Converting division sign into multiplication flips the second fraction.

=\frac{13}{4} \times \frac{12}{7}

Crossing out 12 by 4, we get 3 on the top of second fraction.

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Because problem is about number of rounds.

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Select the correct answer. What is the factored form of 24y5 + 3y2?<br><br>​
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2 years ago
What is the arc measure, in degrees, of major arc \stackrel{\large{\frown}}{ADC} ADC ⌢ A, D, C, start superscript, \frown, end s
Mrrafil [7]

Answer:

\stackrel{\large{\frown}}{ADC}    = 186^{\circ}

Step-by-step explanation:

Given

See attachment

Required

Determine the measure of \stackrel{\large{\frown}}{ADC}

The\ sum\ of\ angles\ in\ a\ circle\ is 360^{\circ}.

So, we have:

\stackrel{\large{\frown}}{ADC} + \stackrel{\large{\frown}}{APB}  + \stackrel{\large{\frown}}{BPC}  = 360^{\circ}

Where:

\stackrel{\large{\frown}}{APB}  = 70^{\circ}

\stackrel{\large{\frown}}{BPC}  = 104^{\circ}

Substitute these values in the above equation.

\stackrel{\large{\frown}}{ADC} + 70^{\circ}   +104^{\circ}   = 360^{\circ}

\stackrel{\large{\frown}}{ADC} + 174^{\circ}   = 360^{\circ}

Collect Like Terms:

\stackrel{\large{\frown}}{ADC}    = 360^{\circ} - 174^{\circ}

\stackrel{\large{\frown}}{ADC}    = 186^{\circ}

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2 years ago
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