What is -0.083 (83 repeating) as a fraction?

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

Mrac [35]

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:

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

So it takes them 5 hours to produce the same amount of widgets.

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

Now we can simply plug in the value of $x=5$ in any of the two equations (i.e. either for Tamara or Jason) to find the number of widgets produced.

$y_{T}=8(5)+20=40+20=60$

So in a time-space of 5 hours Each will have produced 60 widgets and in total will be 60+60 = 120 widgets.

6 0

3 years ago

What is the height of the building pictured below if its shadow is 100 ft. long?

Evgen [1.6K]

Yeah.. baby, do the SOH CAH TOA with the hula hoop

$\bf 
sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
$

so.. hmmm we have in the picture, an angle, and two sides,
actually, is the "adjacent" side, and the "opposite"

so.. which of those ratios give us only
the angle
the adjacent side
the opposite side?

ahhha!, is Ms tangent
so $\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(42^o)=\cfrac{h}{100}$

solve for "h"

5 0

3 years ago

In a series RL circuit, ET = 120 V, R = 40 Ω, and XL = 30 Ω. What is ER?

Delvig [45]

Answer:

The voltage across resistor is $E_{R}=96V$