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Sergeu [11.5K]
3 years ago
7

Sally has a checking account. For each day that Sally's account balance falls below zero, she is charged a fee of $2.50 per day.

Her current balance is -$3.50. If Sally does not make any deposits or withdrawals, what will be the new balance in her account after 4 days?
Mathematics
1 answer:
never [62]3 years ago
5 0

Answer:

-13.50

Step-by-step explanation:

2.50 x -4 = -10

-3.50 + -10 = <u>-13.50</u>

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Given matrices A and B shown below, find (3)A+(8)B.
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Answer:

\left[\begin{array}{ccc}32&-55\\3&39\end{array}\right]

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\left[\begin{array}{ccc}0&-15\\3&15\end{array}\right] +\left[\begin{array}{ccc}32&-40\\0&24\end{array}\right] =\left[\begin{array}{ccc}0+32&-15+(-40)\\3+0&15+24\end{array}\right] = \left[\begin{array}{ccc}32&-55\\3&39\end{array}\right]

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(6x10^2)/(3x10-5) help what is this in standard form?
Tasya [4]

<u>Answer: </u>

The standard form of \frac{6 \times 10^{2}}{3 \times 10^{-5}} is 20,00,0000

<u>Solution: </u>

Given that \frac{6 \times 10^{2}}{3 \times 10^{-5}} ---- eqn 1

To write\frac{6 \times 10^{2}}{3 \times 10^{-5}} in standard form,

We know that \bold{\frac{1}{a^{-m}} = a^{m}} .So \frac{1}{10^{-5}}  becomes 10^{5}.

Now eqn 1 becomes,

= \frac{6 \times 10^{2}}{3} \times 10^{5} ----- eqn 2

We know that \bold{a^{m} \times a^{n}=a^{m+n}}, so 10^{2} \times 10^{5} = 10^{7}

Now eqn 2 becomes,

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Expanding 10^{7}:  

Here 10 is the base term and 7 is the exponent value. So base term 10 is multiplied by itself 7 times.

10^{7} = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10

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= 20,00,0000  

Hence the standard form of \frac{6 \times 10^{2}}{3 \times 10^{-5}} is 20,00,0000

6 0
3 years ago
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