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tangare [24]
3 years ago
11

What is the solution of the system?

Mathematics
1 answer:
Kryger [21]3 years ago
7 0
Greetings!

Solve the System, using Elimination:
\left \{ {{4x+2y=18} \atop {2x+3y=15}} \right.

Multiply Equation #2 by 2:
\left \{ {{4x+2y=18} \atop {2(2x+3y)=2(15)}} \right.

\left \{ {{4x+2y=18} \atop {4x+6y=30}} \right.

Eliminate variable x:
-\frac{ \left \{ {{4x+2y=18} \atop {4x+6y=30}} \right.}{0x-4y=-12}

4y=-12

Divide both sides by 4:
\frac{4y}{4}= \frac{12}{4}

y=3

Input this value into one of the Equations: 
4x+2y=18

4x+2(3)=18

Simplify:
4x+6=18

(4x+6)+(-6)=(18)+(-6)

4x=12

Divide both sides by 4.
\frac{4x}{4}= \frac{12}{4}

x=3

The Solution to this System (The Point of Intersection):
\boxed{(3,3)}

I hope this helped!
-Benjamin
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A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day
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Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

A(t) = A(0)e^{-kt}

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

A(t) = A(0)e^{-0.1392t}

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This is t for which A(t) = 0.5A(0). So

A(t) = A(0)e^{-0.1392t}

0.5A(0) = A(0)e^{-0.1392t}

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\ln{e^{-0.1392t}} = \ln{0.5}

-0.1392t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.1392}

t = 4.98

The substance half-life is of 4.98 days.

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Start by multiplying to get rid of parentheses. Then use addition/subtraction to isolate x on one side of the equation. Finally, use division to determine the value of x.

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What’s the answer????
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Step-by-step explanation:

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