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3 years ago

Help me solve this problem 3 ------ 5+square root 41

1 answer:

3 0

Ya want to rationalize the denomenator
if we mutliply the whole thing by (√41)/(√41), we wil still have a √ there messing things up

we need to remember the difference of 2 perfect squares

remember
(a+b)(a-b)=a²-b²

so, we can see the denomenator is 5+√41
multiply the whole fraction by $\frac{5-\sqrt{41}}{5-\sqrt{41}}$

we get
$\frac{3(5-\sqrt{41})}{5^2-(\sqrt{41})^2}=\frac{15-3\sqrt{41}}{25-41}=$
$\frac{15-3\sqrt{41}}{-16}=\frac{-15+3\sqrt{41}}{16}$

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Tamara and Jason work at a widget factory. Tamara arrived at work before Jason and began making widgets. Tamara had already made

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Answer:

See explanation!

Step-by-step explanation:

The remaining question reads:

Part A: Write a system of equations to represent the situation. Let x = hours and y = widgets.

Part B: How much time does it take for Tamara and Jason to produce the same number of widgets?

Part C: How many widgets will Tamara and Jason have produced?

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To solve this problem we can construct an linear algebraic expression for each person, principally denoted as:

$y=ax+c$

where

$y$ :is our dependent variable (which is a function of $x$ )

$x$ :is our independent variable

$a$ :is the slope

$c$ :is the y-intercept constant value (if any present)

Part A: Write a system of equations to represent the situation. Let x = hours and y = widgets.

Tamara

Has already made 20 widgets and produces at a rate of 8 widgets per hour thus here $c=20$ and $a=8$ , so Tamara's equation reads:

$y_{T}=8x+20$

Jason

Has not produced any widgets yet and produces at a rate of 12 widgets per hour thus here $c=0$ and $a=12$ , so Jason's equation reads:

$y_{J}=12x$

So the system of equations will be

$y_{T}=8x+20\\y_{J}=12x$

Part B: How much time does it take for Tamara and Jason to produce the same number of widgets?

Since we want to find the amount of time (i.e. the value of $x$ ) it takes for both of them to produce the same number of widgets we can just equate the two equations ans solve for $x$ as follow:

$y_{T}=y_{J}\\8x+20=12x\\8x-12x=-20\\-4x=-20\\x=\frac{-20}{-4}\\ x=5$