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andrezito [222]
3 years ago
15

In a different plan for area​ codes, the first digit could be any number from 4 through 8​, the second digit was either 3, 4, 5,

or 6​, and the third digit could be any number exceptnbsp 2 or 5. with this​ plan, how many different area codes are​ possible?
Mathematics
1 answer:
Snezhnost [94]3 years ago
7 0

For the first digit, we have 5 options that are 4,5,6,7,8 . For the second digit, we have 4 options which are 3,4,5 or 6 and for the third digit, we have the options of all numbers except 2 or 5 that is 1,3,4,6,7,8,9,0 . SO we have 8 options for third digit . So to find the total number of options, we need to multiply all the possible options for each digit that is 5 times 4 times 8 = 160 . So the number of possible options are 160 .

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Read 2 more answers
If tan (theta) =5/4 and cos (theta) < 0 . Find the other trig. functions.
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Tan(theta) = 5/4
sin(theta)/cos (theta)=5/4
cos(theta)=4sin(theta)/5
sin²(theta)+cos²(theta)=1
sin²(theta)+16sin²(iheta)/25=1  /*25
25 sin²(theta)+16 sin²(theta)=25
sin²(theta)=25/41    /√
sin(theta)=\frac{-5 \sqrt{41} }{41}
cos(theta)=\frac{-4 \sqrt{41} }{41}
cot(theta)=\frac{4}{5}
csc(theta)=1/sin(theta)=\frac{- \sqrt{41} }{5}
sec(theta)=1/cos(theta)=\frac{- \sqrt{41} }{4}

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