Helen had 5/6 can of orange juice. She spilled 2/6 of a can. How much juice did Helen have left?

Answers

Answer 1
Answer: Given: 5/6 can of orange , 2/6 can of orange spilled

Unknown: How much juice did Helen have left?

Solution: 5/6-2/6=n

Answer: 1/2 juice left for Helen

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How many solutions does the equation 9(x – 3) = 9x – 27 have?

Answers

Answer:2

Step-by-step explanation:

One more question, Will pick brainliest + extra points!! Find x!!

Answers

Here, as the sides are equal, middle line would be exact half then the longer (bottom) side of the triangle. 
48 = 2(3x)
6x = 48
x = 48/6
x = 8

In short, Your Answer would be Option D

Hope this helps!
Since the triangle ADC is an enlargement of triangle AEB, 48 divided by the enlargement factor will give you 3x.
The enlargement factor is:
= 60/30
= 2

Therefore:

3x = 48/2

3x = 24

3x/3 = 24/3

x = 8

0.3(y+2)=1.7+(8-y) solve the following equation​

Answers

Answer:

y = 7

Step-by-step explanation:

0.3y+0.6 = 9.7 - y

0.3y+y = 9.7 -0.6

1.3y = 9.1

y= 7

sorry it had an error

Please help me determine the wedge/dash molecular structure, (R)-5,5-dibromo-3-fluoro-2-methyl-3-hexanol.

Answers

lol this makes absolutly no sense in my mind, what do even half of the words mean? XD

What is the product? assume x>0 (√3x + √5) (√15x +2√30

Answers

The first step for solving this expression is to multiply the parenthesis.
√(45) x^(2) +  2√(90) x +  √(75)x + 2√(150)
Simplify the first radical.
3√(5) x^(2) + 2√(90) x + √(75)x + 2√(150)
Simplify the second radical.
3√(5) x^(2) + 2X3√(10) x + √(75)x + 2√(150)
Simplify the third radical.
3√(5) x^(2) + 2X3√(10) x + 5√(3)x + 2√(150)
Simplify the final radical.
3√(5) x^(2) + 2X3√(10) x + 5√(3)x + 10√(6)
Lastly,, calculate the product of 2 × 3√(10)x to get your final answer.
3√(5) x^(2) + 6√(10) x + 5√(3)x + 10√(6)
Let me know if you have any further questions.
:)

Answer: B: 3x√5+6√10x+5√3x+10√6x

Step-by-step explanation:

Solve for u. -24 = u ÷ 6
U=

Answers

Answer: Um, the answer is U= –144.

Step-by-step explanation: