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4vir4ik [10]
3 years ago
8

At piggys pizza, a giant pizza is 18 inches in diameter. What is the area of this pizza? Round your answer to the nearest tenth

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
7 0
254.5 inches cause its pi times radius squared 
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Compare the functions shown below: g(x) f(x) = −6x − 3 cosine function with y intercept at 0, negative 3 h(x) = 2 cos(x + π) − 1
anastassius [24]

Answer:

because of the product

Step-by-step explanation:

8 0
2 years ago
Indicate whether the measures 7, 7, and 11 can be the side lengths of a triangle. If they can, classify the triangle.
FromTheMoon [43]
Yes they can if 7 + 7 equals more then 11 the. The answer is yes
8 0
3 years ago
3 folders cost $2.91
WARRIOR [948]

Answer: E. None of the above

Step-by-step explanation:

The equation to solve the problem should look like...2x = $2.91 / 3

3 0
2 years ago
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What is the largest shape that can be created by diagonal cut through a square pyramid?
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2 years ago
Radioactive Decay:
Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

<u>Step-by-step explanation:</u>

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}

The formula used to calculate the time period for a first order reaction follows:

t=\frac{2.303}{k}\log \frac{a}{(a-x)}

where,

k = rate constant = 0.01506days^{-1}

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0
2 years ago
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