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Leona [35]
3 years ago
10

S is inversely proportional to t . When s = 0.5 , t = 7 Work out t when s = 5

Mathematics
1 answer:
shtirl [24]3 years ago
4 0

Answer:

t = 0.7

Step-by-step explanation:

s = k/t  where 'k' is the constant of variation

given:  0.5 = k/7 so k = 3.5

5 = 3.5/t

5t = 3.5

t = 0.7

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Answer:

  • 484ed+36_67'ten 355+(36)8wwh
6 0
3 years ago
Pedro has a bag of flour that weighs 9/10 pound. he uses 2/3 of the bag to make gravy. how many pounds of flour does pedro use t
BARSIC [14]
Here is the solution of the given problem above.
First, let's analyze the question. 
Given: 1 bag = 9/10 pound
            2/3 bag = ? pound
What we are going to do is to divide 9/10 pound to 3.
so 9/10 divided by 3 and we get 9/30 and to simplify that, 3/10.
So per 1/3 of the bag, there is 3/10 pound. 
To get the weight for 2/3 of the bag, we multiply 3/10 by 2 and we get 6/10 or to simplify it, it is 3/5. Therefore, the 2/3 bag weights 3/5 pound. Hope this answer helps.
7 0
3 years ago
Read 2 more answers
I WILL GIVE BRAINLIEST!!!
zimovet [89]

Not equivalent to any of the given expressions.

Step-by-step explanation:

{7}^{2m} . {6}^{2m}  \\  \\  = (7.6) ^{2m}  \\  \\  =  {42}^{2m}

Hence, {42}^{2m} is not equivalent to any of the given expressions.

6 0
3 years ago
Read 2 more answers
300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A
Ede4ka [16]

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle +  The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

3 0
3 years ago
-n +(-4) – (-4n) + 6=
Komok [63]

Answer:

3n + 2.

Step-by-step explanation:

Just combine like terms.

4 0
3 years ago
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