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Digiron [165]
3 years ago
9

What is 7/6 reduced to

Mathematics
2 answers:
evablogger [386]3 years ago
8 0
7/6 is 1 and 1/6 (1 1/6) as a reduced fraction. 7/6 is an improper fraction. So, how many times does 6 go into 7, it goes into it one time and there is one remainder.
kotykmax [81]3 years ago
4 0

Answer:

1 1/6

Step-by-step explanation:

7/6 is an improper fraction, so to reduce it, you know that 6/6+1/6=7/6. 6/6=1, and 1/6 is still just 1/6. When you put the two together, you get 1 1/6

You might be interested in
How do i <br> factor 7x^2+3x-4
Bas_tet [7]

Answer:

(x+1) (7x - 4)

Step-by-step explanation:

7x² + 3x - 4

Use the diamond...

7 times -4 equals -28 and 7 plus -4 = 3

Use factor by grouping b/c a does not equal 1

7x² + 7x - 4x - 4

7x(x + 1) -4(x + 1)

(x+1) (7x - 4)

6 0
2 years ago
The number of hours in d days and 6 hours how to write in algebraic expression
kkurt [141]
Let's say the number of days is d (as stated in the question). there are always 24 hours in a day, so the amount of hours in d days has to be 24*d. since you also have the six hours, this can be written as 24*d + 6 = the number of hours in d days and six hours.
6 0
3 years ago
2. Consider the following graph of a quadratic function.
SpyIntel [72]

Answer:

Domain is all real numbers.

Range is -3 to infinity that is y >= -3 (that is y -3)

Increasing from x > -4 ( that is from -4 to +infinity) that is to right of -4

Decreasing from x <-4 (that is -infinity to -4) that is to left of -4

This has a minima at x= -4 with y= -3. We can write as (-4, -3)

Maxima is not there as it is going to infinity

Step-by-step explanation:

5 0
2 years ago
11 student are in a waitlist for a course. two of the students in the list are named ed and ned. how many different ways are the
sesenic [268]

The different ways that are there for the students to be ordered in the list so that Ed is higher on the list than Ned is 19,958,400

In the order, if Ned comes first on the list then Ed will not be on the list before Ned. (remember for Ed to be higher than Ned on the list, it should come after Ned on the list).

So if Ned is first on the list, then all the remaining (including Ed) can occur anywhere in the remaining list. Hence the total permutations will be 10!

Ned's corresponding choices: 1 109 8 7 6 5 4 3 -2 1

If Ned comes in the second position, then Ed can't come in the first position. Hence in the first position, there are only 9 choices.

*(excluding Ed) Ned * corresponding choices:

= 9*9!

*(including Ed (including Ed) Ned corresponding choices: *k

= 9*8*8!

Total combinations will be:

10! + 9*9! + 9*8*8! + 9*8*7*7! + 9*8*7*6*6! + 9*8*7*6*5*5! + 9*8*7*6*5*4*4! + 9*8*7*6*5*4*3*3! +9*8*7*6*5*4*3*2*2! + 9*8*7*6*5*4*3*2*1*1! .

= 19,958,400

We can directly compute it. Since there will be a total of 11! combinations in which 11!/2 times Ed comes after Ned and 11!/2 times Ed comes before Ned.

and 11!/2 = 39,916,800/2 = 19,958,400

To learn more about permutation and combination,

brainly.com/question/1216161

#SPJ4

3 0
1 year ago
Evaluate.<br> X =<br> -(-2)^(-2)2 -4(1)(-8)<br> 2(1)
Vlada [557]

Answer:

x =  \frac{129}{2}

Step-by-step explanation:

x  = ( - 2 {)}^{ - 2}  \times 2 - 4 \times 1 \times ( - 8) \times 2 \times 1

x = ( - 2 {)}^{ - 2}  \times 2 + 4 \times 1 \times 8 \times2  \times 1

x + 2( ( - 2 {)}^{ - 2}  + 4 \times 8)

x = 2( {2}^{ - 2}  + 32)

x =  2( \frac{1}{ {2}^{2} }  + 32)

x = 2( \frac{1}{4}  + 32)

x = 2 \times  \frac{129}{4}

x =  \frac{129}{2}

4 0
3 years ago
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