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Shtirlitz [24]
3 years ago
14

What is 0.666667 as a fraction ?

Mathematics
2 answers:
Olenka [21]3 years ago
7 0
0.666667 as a fraction is 2/3
yanalaym [24]3 years ago
7 0
It is three fifths as .6 would make it 6/10 and when that is reduced it equals 3/5.
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1/3k+80=1/2k+120 what does k equal?
LenaWriter [7]

Answer:

k = -1/240

Step-by-step explanation:

to evaluate the value of k in the expression 1/3k+80=1/2k+120

we have

1/3k+80=1/2k+120

collect the like terms for easy evaluation

1/3k - 1/2k = 120 -80

1/3k - 1/2k = 40

find the lcm

2 - 3/6k = 40

-1/ 6k = 40

cross multiply

6k  x  40  = -1

240k = -1

divide both sides by 240

240k/240 = -1/ 240

k = -1/240

therefore the value of k in the expression 1/3k+80=1/2k+120 is equals to -1/240

8 0
3 years ago
lia gets paid $450 a week pluse commission on het sale. she hopes to earn at least $600 each week. write an inequality to repres
FrozenT [24]
So commission will be $150 450+_=600
3 0
3 years ago
At the Souvenir Shop, gemstone souvenirs cost $11.60 for 5. Which proportion can be used to find the cost of just 1 gemstone?
Ad libitum [116K]
D. If 11.60 of them 5, then in your proportions they would need to be on the same side (both numerators or both denominators) and D is the only answer where this is true.
5 0
3 years ago
A) Work out 4 1/7 + 1 1/2​
Vesnalui [34]

Answer:

135/14

9 9/14

Step-by-step explanation:

29/7+11/2

4 0
2 years ago
Read 2 more answers
Will a pair of 6 feet 11 inch long skis fit in a wardrobe that is 3 feet long, 2 feet wide and 6 feet high?
JulsSmile [24]

Answer:

Step-by-step explanation:

Length of a pair of skis = 6 feet 11 inches

Maximum length of a the skis that can be fitted in a wardrobe is along the diagonal of the wardrobe.

From the picture attached,

BC² = 3² + 2² [By Pythagoras theorem]

BC = √(13)

By applying Pythagoras theorem in ΔABC,

AB² = AC² + BC²

AB² = 6² + (√13)²

       = 36 + 13

AB = √49

Ab = 7 feet

Since, "diagonal length of the wardrobe is more than the length of skis,

skis can be fitted"

3 0
3 years ago
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