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Agata [3.3K]
2 years ago
7

PLEASE PLEASE HELP

Mathematics
1 answer:
marta [7]2 years ago
3 0

Answer:

Approximately 3 grams left.

Step-by-step explanation:

We will utilize the standard form of an exponential function, given by:

f(t)=a(r)^t

In the case of half-life, our rate <em>r</em> will be 1/2. This is because 1/2 or 50% will be left after <em>t </em>half-lives.

Our initial amount <em>a </em>is 185 grams.

So, by substitution, we have:

\displaystyle f(t)=185\big(\frac{1}{2}\big)^t

Where <em>f(t)</em> denotes the amount of grams left after <em>t</em> half-lives.

We want to find the amount left after 6 half-lives. Therefore, <em>t </em>= 6. Then using our function, we acquire:

\displaystyle f(6)=185\big(\frac{1}{2}\big)^6

Evaluate:

\displaystyle f(6)=185\big(\frac{1}{64}\big)\approx2.89\approx 3

So, after six half-lives, there will be approximately 3 grams left.

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Which expression is equivalent to (5x^5)(4x)^3​
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5x. 5x. 5x. 5x. 5x. 4x. 4x. 4x

Step-by-step explanation:

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3 years ago
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

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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

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3 0
3 years ago
-8(8v + 1) - 2 = -394
Bess [88]

Answer:

v = 6

Step-by-step explanation:

Solve for v:

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Grouping like terms, -64 v - 8 - 2 = -64 v + (-8 - 2):

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