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AfilCa [17]
3 years ago
7

Ren was trying to solve a system of linear equations. He correctly argued that he could replace one equation by the sum of that

equation and a multiple of the other. H
produce a system that has the same solution as the original system.
To show this, Ren wrote down a system of equations. He then determined which other systems of equations would have the same solution as his original system. Rem
below.
-2x - 5y = -47
4x - 2y = -2
Complete the table below to determine which system(s) of equations has the same solution as the original system shown above. Choose Has the Same Solution or Do
for each system of equations.
Has the Same Solution Does Not Have the Same Solution
-14x + y = -41
4x - 2y = -2
-2x - 5y = -47
6x + 3y = 45
-2x - 5y = 47
8x - 5y = -8
0 0 0
Mathematics
1 answer:
Fittoniya [83]3 years ago
7 0

Answer:

1. yes 2. yes 3. no

Step-by-step explanation:

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What is36.8+[11.6-(2.5×3)]2
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Answer:

Step-by-step explanation:

The answer is 33.4

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A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest
faltersainse [42]
\bf \qquad \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 &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
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\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

\bf \qquad \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 &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396

add both amounts, and that's her investment for the 11 years.
7 0
3 years ago
I need help with the top 3 plz
attashe74 [19]

Answer/Step-by-step explanation:

Recall: SOHCAHTOA

1. Reference angle = 70°

Adjacent side = x

Hypotenuse = 6 cm

Apply CAH. Thus,

Cos 70 = adj/hyp

Cos 70 = x/6

6 × cos 70 = x

2.05 = x

x = 2.05 cm

2. Reference angle = 45°

Adjacent side = x

Hypotenuse = 1.3 m

Applying CAH, we would have the following ratio:

Cos 45 = adj/hyp

Cos 45 = x/1.3

1.3 × cos 45 = x

0.92 = x

x = 0.92 m

3. The who diagram is not shown well. Some parts are missing, however you can still solve the problem just the same way we solved problem 1 and 2.

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