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3 years ago

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An investor has an account with stock from two different companies. Last year, his stock in Company A was worth $1790 and his st

ock in Company B was worth $1000. The stock in Company A has increased 20% since last year and the stock in Company B has increased 22%. What was the total percentage increase in the investor's stock account? Round your answer to the nearest tenth (if necessary).

2 answers:

uysha [10]3 years ago

8 0

Answer:

20.7%

Step-by-step explanation:

stiv31 [10]3 years ago

7 0

Answer:

20.7%

Step-by-step explanation:

Amount of stock increased in Company A = 20% of 1790

= 0.20 * 1790 = $ 358

Investor's stock in company A this year = 1790 + 358 = $ 2148

Amount of stock increased in Company B = 22% of 1000

= 0.22 * 1000= $ 220

Investor's stock in company B this year = 1000 + 220 = $ 1220

Total amount in investor's stock account = 2148 + 1220 = $ 3368

Increased amount = 3368 - 2790 = $ 578

Increase percentage = $\frac{578}{2790}*100$ = 20.71

= 20.7%

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Solve the system of equations by using the inverse of the coefficient matrix of the equivalent matrix equation.

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The solution for the given system of equations x + 8y = -37, 4x + 8y = -52 is $\left[\begin{array}{ccc}&5&\\\\&4&\end{array}\right]$

Given,

System of equations as,

x + 8y = -37

4x + 8y = -52

We have to solve this by using the inverse of coefficient matrix of the equivalent matrix equation.

That is,

$A=\left[\begin{array}{ccc}a&&b\\\\c&&d\end{array}\right]$

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Now we can solve the equations.

Here we have,

x + 8y = -37

4x + 8y = -52

Now in matrix form,

$\left[\begin{array}{ccc}1&&8\\\\4&&8\end{array}\right]$ $\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right]$ $=\left[\begin{array}{ccc}&-37&\\\\&-52&\end{array}\right]$

A X B

We know that,

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Therefore,

$A^{-1} = \frac{1}{(1X8)-(4X8)} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{8-32} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{-24} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$

Then,

$\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right] =$ $=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$ $\left[\begin{array}{ccc}&\frac{37}{52} &\\\\\end{array}\right]$

$=\frac{1}{24} \left[\begin{array}{ccc}-8&&8\\\\4&&-1\end{array}\right]$ $\left[\begin{array}{ccc}&\frac{37}{52} &\\\\\end{array}\right]$

$=\frac{1}{24} \left[\begin{array}{ccc}(-8X37)+(8X52)\\\\(4X37)+(-1X52)\end{array}\right]$

$=\frac{1}{24} \left[\begin{array}{ccc}-296+416\\\\148-52\end{array}\right]$

$=\frac{1}{24} \left[\begin{array}{ccc}120\\\\96\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{120}{24} \\\\\frac{96}{24} \end{array}\right]$

$=\left[\begin{array}{ccc}5\\\\4\end{array}\right]$

That is $\left[\begin{array}{ccc}x\\\\y\end{array}\right] =\left[\begin{array}{ccc}5\\\\4\end{array}\right]$

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The question is incomplete. Completed question is given below:

Solve The System Of Equations By Using The Inverse Of The Coefficient Matrix Of The Equivalent Matrix Equation.

x + 8y = -37

4x + 8y = -52

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2 years ago

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Answer:

Step-by-step explanation:

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$A=A_0(\frac{1}{2})^{\frac{t}{H}$ where A is the amount after some decay has happened, A₀ is the intial amount, t is the time in hours, and H is the half-life in hours. Those values for us are:

A = 2 g

A₀ = 25 g

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t = ?

Filling in:

$2=25(\frac{1}{2})^{\frac{t}{4.95}$ Keep in mind that, because of the nature of the exponential form of this equation, you CANNOT simply multiply the 25 by the 1/2. Exponential equations don't work that way. Begin instead by dividing both sides by 25 to get

$.08=(\frac{1}{2})^{\frac{t}{4.95}$ The goal is to get that t out from its exponential position. Do that by taking the natural log of both sides:

$ln(.08)=ln(.5)^{\frac{t}{4.95}$

After you take the natural log of the right side, the property allows you to bring the exponent down out front:

$ln(.08)=\frac{t}{4.95}ln(.5)$

Now divide both sides by ln(.5) to get

$\frac{ln(.08)}{ln(.5)}=\frac{t}{4.95}$

Simplify the left side out on your calculator to get

$\frac{-2.525728644}{-.6931471806}=\frac{t}{4.95}$ and then divide:

$3.643856=\frac{t}{4.95}$

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3.643856(4.95) = t so

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4 years ago