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ANTONII [103]
3 years ago
12

Select the correct structure that

Mathematics
2 answers:
vampirchik [111]3 years ago
6 0

Answer: B

Step-by-step explanation:

Veseljchak [2.6K]3 years ago
5 0

Answer:

the correct structure would be B

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Julie's overtime pay is $7 an hour more than her regular pay. She worked 6 hours at her regular wage and 4 hours at her overtime
sattari [20]
--------------------------------------
Define x:
--------------------------------------
Let her regular hourly pay be x.
Overtime pay = x + 7

--------------------------------------
Construct equation:
--------------------------------------
6x + 4(x+ 7) = 148
6x + 4x + 28 = 148
10x = 148 - 28
10x = 120
x = 12

--------------------------------------
Find Regular Pay
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Regular Pay = x = $12

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Answer: Her regular pay is $12 an hour.
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7 0
4 years ago
How to calculate how many times greater a number is?
stiv31 [10]
Division is ur answer..

examples...

300 and 30......300/30 = 10....so 300 is 10 times greater then 30.

5000 and 50....5000/50 = 100...so 5000 is 100 times greater then 50.

70 and 20 ......70/20 = 3.5 ....so 70 is 3.5 times bigger then 20


5 0
3 years ago
For what values of theta(0
Art [367]

Answer:

pi/2

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Four angles of a pentagon measures 123, 64, 171 and 77. Find the measure of the fifth angle.
cricket20 [7]

Answer:

69

Step-by-step explanation:

4 0
3 years ago
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An algorithm takes 0.5 seconds to run on an input of size 100. How long will it take to run on an input of size 1000 if the algo
dexar [7]

Answer:

linear: 5s

quadratic: 50s

log-linear: 0.75 s

cubic: 500s

Step-by-step explanation:

Let t_1,t_2 be the running time associated with the input of sizes s_1,s_2

If the running time is linear

t_2 = t_1\frac{s_2}{s_1} = 0.5*\frac{1000}{100} = 0.5*10 = 5s

If the running time is quadratic

t_2 = t_1\left(\frac{s_2}{s_1}\right)^2 = 0.5*\left(\frac{1000}{100}\right)^2 = 0.5*10^2 = 50s

If the running time is log-linear

t_2 = t_1\frac{log(s_2)}{log(s_1)} = 0.5*\frac{log(1000)}{log(100)} = 0.5*1.5 = 0.75s

If the running time is cubic:

t_2 = t_1\left(\frac{s_2}{s_1}\right)^3 = 0.5*\left(\frac{1000}{100}\right)^3 = 0.5*10^3 = 500s

6 0
3 years ago
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