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lukranit [14]
4 years ago
9

The sum of two numbers is 53 and the difference is 15 . What are the numbers?

Mathematics
2 answers:
Stells [14]4 years ago
6 0

Answer:

34 and 19.

Step-by-step explanation:

Since the sum of the two numbers is 53, x+y=53. Since the difference is 15, x-y=15.

Adding x+y=53

and       x-y=15,

The ys cancel, leaving 2x=68, or x=34. Therefore, 34-y=15, or y=19.

cluponka [151]4 years ago
3 0

Answer:

34 and 19.

Step-by-step explanation:

If the numbers are x and y:

x + y = 53

x - y = 15       Add the 2 equations:

2x + y - y = 68

2x = 68

x = 34.

As they add up to 53:

y = 53 - 34 = 19.

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Suppose the average of 15 consectutive numbers is 15. What is the average of the first five numbers of the set?
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7 0
2 years ago
6/10+6/100= how do i do this problem
olya-2409 [2.1K]

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Step-by-step explanation:

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8 0
3 years ago
You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp
Katena32 [7]
Let's move like a crab, backwards some.

after 2 years?

\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 &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2

after 3 years?

\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 &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3

is that enough to pay the $1100?


now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000   <---- standard form.
8 0
3 years ago
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