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bearhunter [10]
3 years ago
8

PLEASE HELP ME ASAP!! THANK YOU Write each of the following numbers as a power of the number 2:

Mathematics
1 answer:
DaniilM [7]3 years ago
6 0

Answer:

32 = 2^5 and 16=2^4.......

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What is the greatest common factor of 16ab3 + 4a2b + 8ab ?
ziro4ka [17]

Answer:

2ab(3b^2+2a+4)

Step-by-step explanation:

6ab^3 + 4a^2b + 8ab

2*3*a*b*b^2 +2*2*a*a*b +2*2*2*a*b

Factor out the common terms

2ab( 3*b^2 +2*a +2*2)

2ab(3b^2+2a+4)

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3 years ago
A car takes 54.9 seconds to travel 1 mile. How long does it take the car to travel 4.3 miles? First round to the nearest whole n
fiasKO [112]

Answer: Rounded: 236 seconds

Not rounded: 236.07 seconds

Step-by-step explanation: We know that it takes the car 54.9 seconds to travel one mile, and we know that 1 times 4.3 is 4.3, so 4.3 times 54.9 equals 236.07 seconds.

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3 years ago
What is a gradient of a line
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(slope) The higher the gradient<span> of a graph at a point, the steeper the </span>line<span> is at that point.</span>
4 0
3 years ago
Read 2 more answers
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

^nC_r=\dfrac{n!}{(n-r)!r!}

Given : Total multiple-choice questions  = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= ^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84

No. of ways to answer 4 open-ended problems

= ^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= 84\times15=1260

Hence, the correct answer is option A) 1260

5 0
3 years ago
10. How many three-topping pizzas can be ordered from the list of toppings below? Did you calculate the number of permutations o
Nimfa-mama [501]

Step-by-step explanation:

Total numbers of pizza toppings = 12

Number of three topping pizzas can be ordered are:

P^{n}_{k}=\frac{n!}{(n-k)!}

where = n = number of elements

k = number of elements choosen

n= 12 , k = 3

\frac{12!}{(12-3)!} =\frac{12\times 11\times 10\times 9!}{9!}= 1,320

We calculated the number of permutations. We made this choice because in permutation order of element matters but in combination its not.

8 0
3 years ago
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