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3 years ago

Simplify all the way

Mathematics

1 answer:

attashe74 [19]3 years ago

4 0

Answer:

Step-by-step explanation:

3^7/3^3

3*3*3*3*3*3*3=2187

3*3*3=27

2187/27=81

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I put $500 in a saving account that warns 7% interest compounded quarterly. How much will be in the account after 1 year? Rounde

Fantom [35]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 7\%\to \frac{7}{100}\to &0.07\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &1
\end{cases}
\\\\\\
A=500\left(1+\frac{0.07}{4}\right)^{4\cdot 1}\implies A=500(1.0175)^4\implies A\approx 535.92951564$

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3 years ago

What is the answer to 0.1 * n = 89.9 ?

MrRissso [65]

Divide both sides of the equation by 0.1 which leaves you with
n = 89.9/0.1 = 899

4 0

3 years ago

Read 2 more answers

Simplify each expression.

Vikki [24]

Answer:

is number 3

Step-by-step explanation:

see explanation on the attached

hope it helps

8 0

3 years ago

Maggie needs to spend at least six hours each week practicing the piano. She has already practiced three and one fourth hours th

Allisa [31]

Answer:

3 1/4 + 2x ≥ 6

Step-by-step explanation:

Let X equal the remaining time she needs to practice.

You would have 2x

The combined total needs to be 6 hours or greater.

You need to add the amount she already practiced to 2x.

Now you have: three and one fourth + 2x

This needs to be greater than or equal to 6.

I hope this helps you :)

6 0

3 years ago

Product of two natural numbers p and q is 590 and their hcf is 59. how many set of values of p and q is possible

djverab [1.8K]

$\text{Let the product of two natural numbers p and q is 590, and their HCF is 59}\\
\\
\text{we know that the product of LCM and HCF of any two numbers is equal}\\
\text{to the product of the numbers. that is}\\
\\
\text{HCF}\times \text{ LCM}=p\times q\\
\\
\Rightarrow 59 \times \text{LCM}=590\\
\\
\Rightarrow \text{LCM}=\frac{590}{59}\\
\\
\Rightarrow \text{LCM}=10\\
\\
\text{for any two natural numbers, their Least Common Multiple (LCM) is always}$

$\text{greater than their HCF.}\\
\\
\text{but here we can see that }LCM$

Hence there is no such natural numbers exist.

7 0

3 years ago