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AveGali [126]
4 years ago
10

How can x2+3x+1=2x2+2x+3 be set up as a system of equations

Mathematics
1 answer:
Vesna [10]4 years ago
6 0
For this case we can write the system of equations as two equations with two unknowns.
 We then define the following unknowns:
 incognitas: x and y.
 We write the system of equations:
 y = x2 + 3x + 1
 y = 2x2 + 2x + 3
 Answer:
 
y = x2 + 3x + 1
 
y = 2x2 + 2x + 3
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Hope I helped!!


~A


3 0
3 years ago
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Lin's Job pays $8.25 an hour plus $10 of transportation allowance ench week. She has to work at least 5 hours a work to keep the
marshall27 [118]

Answer:

x ≥ 5, 8.25x + 10 = 175, and x = 20 hours.

Step-by-step explanation:

Let Lin works x hours per week.

Now given that Lin’s job pays $8.25 an hour plus $10 of transportation allowance each week.

And she has to work at least 5 hours a week to $175 per week.

So, the inequality that can be written from the above condition is x ≥ 5 .......(1)

And the equation that can be written that  

8.25x + 10 = 175 ....... (2)  

Now, fro equation (2) we can write x = 20 hours which satisfies equation (1).

5 0
3 years ago
Can someone help me with this math homework please!
labwork [276]

Answer:

1/2

8

Step-by-step explanation:

When its talking about the result or output, look at the range, and then follow the line(s) back to the number(s) in the domain. Do the opposite when it's talking about the function of a certain number, e.g. f(4).

7 0
3 years ago
A coat and a pair of boots are on sale for 20% off. The regular price of the coat is $225. The regular price of the boots is $13
kondaur [170]

Answer:

Part A = $180

Part B = $104

Step-by-step explanation:

Part A:

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Part B:

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7 0
3 years ago
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226Ra has a half-life of 1599 years. How much is left after 1000 years if the initial amount was 10 g?
Ray Of Light [21]
\bf \textit{Amount for Exponential Decay using Half-Life}\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &10\\
t=\textit{elapsed time}\to &1000\\
h=\textit{half-life}\to &1599
\end{cases}
\\\\\\
A=10\left( \frac{1}{2} \right)^{\frac{1000}{1599}}
8 0
3 years ago
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