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andrezito [222]
3 years ago
15

Please need the answer as fast as possible! 1/6 + 2/3​

Mathematics
2 answers:
laiz [17]3 years ago
7 0

Answer:

5/6

Step-by-step explanation:

I hope this helps you you gotta convert the 2/3 into 4/6 and then you add 4/6 + 1/6= 5/6

MrMuchimi3 years ago
5 0
The least common multiple of six and three is six multiply the numerator and denominator of each fraction by whatever value will result and the denominator of each fraction being equal to the least common factor which is six.
When they have like denominators which is the same ones you can add
1/6+2/3
1/6+2/3 times 2 by 2 and 3 by 2 with the 2/3 fraction
Then u get 1/6+4/6
=1+4/6
=5/6
Mark brainliest if I helped/answer correct !
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Solve the equation for x.​
Ghella [55]

Answer:

- x/10 = 9

multiply by 10 on both sides

-x = 90

divided by -1 on both sides

x = -90

6 0
3 years ago
Find the equation of the line.Use exact numbers.<br><br> y= ___ x+ ___
Alona [7]
Answer: y=3x+3

explanation:
7 0
3 years ago
I need help please thanks
konstantin123 [22]

Answer:

1/6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What transformation is represented by the rule (x, y)→(y, − x) ?
Fofino [41]
The first on is: <span>rotation of 90° clockwise about the origin

the second one should be: </span><span>a reflection over y = x 

and the third one is: </span><span>a translation of 3 units to the left 

hope this helps.........</span>
8 0
3 years ago
Read 2 more answers
The floor of a conference hall can be covered completly with tiles.Its length is 36ft. longer than its width.The area of the flo
Masteriza [31]

Answer:

Step-by-step explanation:

Given

Area of a floor = 2040ft²

If Its length is 36ft. longer than its width, then L = W+36

Area = LW

L is the length

W is the width

A <  W(W+36)

2040 < W²+36W

0<W²+36W-2040

W²+36W-2040 >0

W = -36±√36²-4(-2040)/2

W =  -36±√1296+8160)/2

W =  -36±√9456/2

W =  -36±97.25/2

W = -36+97.25/2

W = 61.24/2

W > 30.62ft

Since L = W+36

L > 30.62+36

L > 66.62ft

The width of the floor is 30.62ft

The length is 66.62ft

The mathematical sentence would represent the given situation is the width is greater than 30.62ft while the length must be greater than 66.62ft

The possible dimension of the floor is 31ft by 67 ft

The possible areas is 31*67 = 2077ft²

It won't be realistic to get an area of 144sqft because the initial area is greater than 144. Hence the feasible area will be greater than 2040ft²

5 0
3 years ago
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