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3 years ago

Which of the following stanements shows the inverse property of addition?

Mathematics

1 answer:

Nikitich [7]3 years ago

4 0

Answer:

$0\cdot \:y+\left(-0\cdot \:y\right)=0$ the statement shows the inverse property of addition as the sum of a number and its inverse is zero.

Step-by-step explanation:

As we know that Inverse Property of Addition states if you add any number to its opposite, the result will be zero.

For example, 13 + (-13) = 0 shows that -13 is the additive inverse of 13.

In other words, the sum of a number and its inverse is zero.

Now checking from the available options,

$0y+\left(-0y\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=0\cdot \:y-0\cdot \:y$

$\mathrm{Add\:similar\:elements:}\:0\cdot \:y-0\cdot \:y=0$

$=0$

Thus,

$0\cdot \:y+\left(-0\cdot \:y\right)=0$

Therefore, $0\cdot \:y+\left(-0\cdot \:y\right)=0$ the statement shows the inverse property of addition as the sum of a number and its inverse is zero.

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3 years ago

You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which

stellarik [79]

To model this situation, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial deposit
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

For account A:
We know for our problem that $P=2000$ and $r= \frac{2.25}{100} =0.0225$ . Since the interest is compounded monthly, it is compounded 12 times per year; therefore, $n=12$ . Lets replace those values in our formula:
$A=2000(1+ \frac{0.0225}{12} )^{12t}$

For account B:
$P=2000$ , $r= \frac{3}{100} =0.03$ , $n=12$ . Lest replace those values in our formula:
$A=2000(1+ \frac{0.03}{12} )^{12t}$

Since we want to find the time, $t$ , <span>when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span> $2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000$

Now that we have our equation, we just need to solve for $t$ :
$2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000$
$(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}$
$(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}
{2}$
$ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})$
$12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})$
$t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})$
$t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}$
$17.47$

We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>

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3 years ago