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Neporo4naja [7]
3 years ago
11

Can someone help me I’m confused#5

Mathematics
2 answers:
Pani-rosa [81]3 years ago
5 0
I think your supposed to multiply 4 times 3
MrRissso [65]3 years ago
4 0

Answer:

Multiply

Step-by-step explanation:

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The beginning checkbook balance of Gregory Co. was $3,045.58. Their bank statement indicated a balance of $4,262.92. The bookkee
Alex_Xolod [135]

Answer:

$3,659.82

Step-by-step explanation:

Adding to the checkbook those debits and credits not already there brings its balance to ...

... $3045.58 +651.84 -37.60 = $3659.82

Adjusting the bank's balance by the deposits and checks not already there brings its balance to ...

... $4262.92 +220.05 -325.50 -497.65 = $3659.82

Thus, the reconciled accounts will agree on the balance $3659.82.

8 0
3 years ago
What is a way to write 11 more than twice x is less than 44
madam [21]

(2x) + 11 < 44

x is multiplied by two because we are looking for twice of x. 11 is added to the 2x because we are looking for eleven more than the sum of twice of x. Since 2x + 11 is less than 44, we use the lesser than ( < ) symbol.

4 0
3 years ago
Read 2 more answers
4x+1+8-x+5x-2=23 linear equations
DanielleElmas [232]

Step-by-step explanation:

= 4x-x+5x+1+8-2

=8x+7

proved##

5 0
3 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
3 years ago
1.In triangle TRS, VZ = 6 inches. What is RZ?
Hunter-Best [27]
1. C
2. B,C,E

Hope that helps.
4 0
3 years ago
Read 2 more answers
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