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LiRa [457]
3 years ago
6

Can someone please help me?

Mathematics
2 answers:
kolezko [41]3 years ago
6 0
The answer is a where are the other questions?
docker41 [41]3 years ago
4 0
THe answer is A.... 42.4 times 10.5 is 445.20
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Solve for c<br><br><br> R=5(c/a-0.3)
aalyn [17]

we have

R=5(c/a-0.3)

Solve for c--------> that means that clear variable c

so

Divide by 5 both sides

R/5=(c/a-0.3)

Adds 0.3 both sides

(R/5)+0.3=(c/a)

Multiply by a both sides

a[(R/5)+0.3]=c

so

c=a[(R/5)+0.3]

therefore

<u>the answer is</u>

c=a[(R/5)+0.3]

6 0
3 years ago
Read 2 more answers
Jackson wrote the equation below on the whiteboard.
faust18 [17]
B = 7

7 x 6 = 42

42 + 6 = 48.
6 0
3 years ago
Read 2 more answers
2 7/8 plus 3 2/8 plus 1 1/8
Olin [163]

Answer:

7.25

Step-by-step explanation:

2 7/8 + 3 2/8 * 1 1/8 = 7.25

7 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
a portion of 7200 is invested at 5 1/2 percent interest, and the rest is invested at 5 percent interest. if the yearly income fr
lana [24]

Answer:

A 5.5% = 5400 B 5% = 1800

Step-by-step explanation:

A + B = 7200

A * 0.055 + B * 0.05 = $387

A = 7200 - B

(7200 - B)*0.055 + B*0.05 = 387

396 - 0.055B + 0.05B = 387

396 - 0.005B = 387

0.005B = 9

B = 1800

A = 7200 - B = 7200 - 1800

A = 5400

5 0
3 years ago
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