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yulyashka [42]
3 years ago
11

6. Sec(300)= I’m pretty certain this is undefined.

Mathematics
1 answer:
bekas [8.4K]3 years ago
3 0

\sec 300=\dfrac{1}{\cos 300}=\dfrac{1}{\dfrac{1}{2}}=2

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The county fair chargers $1.25 per ticket for the rides. Jermaine bought 25 tickets for the rides and spent a total of $43.75 at
zlopas [31]

let's see what to do buddy....

The county fair charges 1.25 $ per ticket:

Stop :

So if Jermain has bought 25 tickets, he has already spent that much money which equal :

25 \times 1.25 = 25 \times 125 \times  {10}^{ - 2} = 25 \times (100 + 25) \times  {10}^{ - 2}  \\  = (2500 + 625) \times  {10}^{ - 2} = 3125 \times  {10}^{ - 2} = 31.25

So jermain spent 31.25 dollors only for the tickets.

But he spent the total of 43.75 $.

Look : 43.75 $ - 31.25 $ = 12.5 $

Where are the 12.5 $ spent ?

Easy ----¢ This is the cost of fair admission.

Due of all those things we have to pay 1.25 $ per ticket and 12.5 $ for fair admission.

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A) <em><u>x</u></em> as the number of the ride tickets

<em><u>y</u></em> as the total cost we should pay

or as the total money we should

spend

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B)

y = 1.25 \: x + 12.5

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C) I already did buddy.

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5 0
3 years ago
Need help on angles of polygons please
aleksandrvk [35]

Answer:

x+87+92+105+135=540(sum of interior angle of pentagon)

x+419=540

x=540-419

x=121

Step-by-step explanation:

4 0
3 years ago
I really need help ! i’m failing math<br> i need the answers x and y
Elina [12.6K]

Answer:

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3 years ago
A 10-year study (lasting from January 1, 2008 to December 31, 2017) was conducted examining the risk of developing lung cancer a
Aleks [24]

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Detailed step wise solution is given below:

6 0
4 years ago
Maria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there we
alekssr [168]

Answer: 8,800,000

Step-by-step explanation:

First year recorded number of cars on the roads   = 80, 000,000

Increase rate                                                              = 10%

In the second year , the number of cars recorded  = 10% of 80,000,000

                                                                                    = 10/100 x 80,000,000

                                                                                    = 8,000,000

                                                                                =  80,000,000 + 8,000,000

                                                                                    = 88,000,000

In the third year, the number of cars recorded       = 10% of 88,000,000

                                                                                   = 10/100 x 88,000,000

                                                                                   = 8,800,000

                                                                                  = 88,000,000 + 8,800,000

                                                                                  = 96,800,000.

The number of cars we had  more in the third years compared to year two will be

                                                                                = 96,800,000 - 88,000,000

                                                                                = 8,800,000

           

6 0
4 years ago
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