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g100num [7]
2 years ago
10

Nancy’s grandma needs 5/8

Mathematics
1 answer:
Gre4nikov [31]2 years ago
6 0

Answer:16

Step-by-step explanation:

10÷5/8=16

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Please answer this correctly
Tanya [424]

Answer:

All make numbers make that statement true

Step-by-step explanation:

6406358 + 8550780= 14957138

14957138-5598256= 9358882

therefore is you add any number to 5598256 that is greater than 9358882 then the statement will be true. All those numbers given are larger than 9358882

6 0
3 years ago
(4x + 3) (3x^2 - 5x + 6)
Liula [17]

Answer:

12x^3 - 11x^2 + 9x + 18

Step-by-step explanation:

12x^3 - 20x^2 + 24x + 9x^2 -15x + 18

Combine like terms

12x^3 - 11x^2 + 9x + 18

4 0
3 years ago
What is the expanded form of -3/8 (y-x) =
Aloiza [94]

Answer:

look it up  its the same as the equation

Step-by-step explanation:

7 0
3 years ago
shanes neighbor pledged 1.25 for every 0.5 miles that shane swims in the charity swim-a-thon.if Shane swims 3 miles,how much mon
Paraphin [41]
3/0.5 = 6
6 x 1.25 = 7.50
4 0
3 years ago
Read 2 more answers
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
3 years ago
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