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monitta
3 years ago
13

Eric has a gift card to his favorite restaurant. Every Tuesday he has the same meal at the restaurant. This table shows the bala

nce of money in dollars left on Eric’s gift card at different times. Time (weeks) 2 4 6 8 Balance ($) 125 100 75 50 What is the equation that represents Eric’s gift card balance, where x represents the time and y represents the balance? Enter your answers in the boxes. y−100= (x− )
Mathematics
1 answer:
Helen [10]3 years ago
4 0

<u>Answer-</u>

y-100=-\frac{25}{2}(x-4)

<u>Solution-</u>

We can generate an equation that will represent Eric’s gift card balance, by taking two points from the table and applying two point formula to get the linear relationship between card balance and time.

Taking the time on abscissa and card balance on ordinate, the two points would be (2, 125), (4, 100)

\Rightarrow \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}

\Rightarrow \frac{y-125}{100-125}=\frac{x-2}{4-2}

\Rightarrow \frac{y-125}{-25}=\frac{x-2}{2}

\Rightarrow -\frac{y}{25}+5=\frac{x}{2}-1

\Rightarrow \frac{x}{2}+\frac{y}{25}-6=0

\Rightarrow \frac{25x+2y-300}{50}=0

\Rightarrow 25x+2y-300=0

\Rightarrow y=\frac{300-25x}{2}

\Rightarrow y=150-\frac{25}{2}x

This is the equation for the Eric’s gift card balance.

Comparing this equation with general slope-intercept equation of the line, we get

Slope=-\frac{25}{2} \ and\ y-intercept=150

Applying point-slope formula taking point as (4, 100), we get

\Rightarrow y-y_1=m(x-x_1)

\Rightarrow y-100=-\frac{25}{2}(x-4)

This is the desired equation.


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1. To solve this we are going to find the distance of the three sides of each triangle, and then, we are going to use Heron's formula.
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S (2,1), T (1,3) and R (5,5). Using a graphic tool or the distance formula d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} we realize that ST=2.2, SR=5, and TR=4.5. Now we are going to find the semi-perimeter of our triangle:
s= \frac{ST+SR+TR}{2}
s= \frac{2.2+5+4.5}{2}
s= \frac{11.7}{2}
s=5.85
Now we can use Heron's formula:
A= \sqrt{s(s-ST)(s-SR)(s-TR)}
A= \sqrt{5.85(5.85-2.2)(5.85-5)(5.85-4.5)}
A=4.9 square units
We can conclude that the area of triangle RTS is 4.9 square units.

- For triangle MNL:
M (2,-4), N (-2,-3), and L (0,-1). Once again, using a graphic tool or the distance formula we get that MN=4.1, ML=3.6, and NL=2.8.
Lets find the semi-perimeter of our triangle to use Heron's formula:
s= \frac{4.1+3.6+2.8}{2}
s=5.25
A= \sqrt{5.25(5.25-4.1)(5.25-3.6)(5.25-2.8)}
A=4.9 square units
We can conclude that the area of triangle MNL is 4.9 square units.

2. To find the volume of our pyramid, we are going to use the formula for the volume of a rectangular pyramid: V= \frac{lwh}{3}
where
l is the length of the rectangular base
w is the width of the rectangular base 
h is the height of the pyramid
From our picture we can infer that l=15, w=10, and h=12, so lets replace the values in our formula:
V= \frac{lwh}{3}
V= \frac{(15)(10)(12)}{3}
V=600 cubic units 

To find the volume of the cone, we are going to use the formula: V= \pi r^2 \frac{h}{3}
where
r is the radius 
h is the height 
From our picture we can infer that the diameter of the cone is 9; since radius is half the diameter, r= \frac{9}{2} =4.5. We also know that h=12, so lets replace the values:
V= \pi r^2 \frac{h}{3}
V= \pi (4.5)^2 \frac{12}{3}
V=254.5 cubic units 
We can conclude that the volume of the pyramid is 600 cubic units and the volume of the cone is 254.5 cubic units 

3. To find y, we are going to use the tangent trigonometric function:
tan( \alpha )= \frac{opposite.side}{adjacent.side}
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y= \frac{4cm}{tan(53)}
y=3.01cm
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y=3cm
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V=(12)^3
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V= \frac{4}{3} \pi r^3
V= \frac{4}{3} \pi (6)^3
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x+2x+x=180
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x= \frac{180}{4}
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3 0
3 years ago
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