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inna [77]
3 years ago
12

The crew on a fishing boat caught 4 fish that weighed a total of 1,092 pound. The tarpon weighed twice as much as the amberjack

and the white marlin weighed twice as much as the tarpon. The weight of the tuna was 5 times the weight of the Amber jack. How much did each fish weigh?
Mathematics
1 answer:
Savatey [412]3 years ago
7 0

ddddd

Let T be the weight of Tarpon

Let A be the weight of amberjack

Let W be the weight of white marlin

Let TU be the weight of Tuna

The crew on a fishing boat caught 4 fish that weighed a total of 1,092 pound.

Total weight = 1092

T + A + W + TU = 1092

The tarpon weighed twice as much as the amberjack

T= 2 A

  the white marlin weighed twice as much as the tarpon.

W = 2 T

We know T is 2A  so W = 2(2A) = 4 A, W = 4A

The weight of the tuna was 5 times the weight of the Amber jack.

TU = 5A

T = 2A  , W = 4A , TU = 5A

Replace it in our equation

T + A + W + TU = 1092

2A + A + 4A + 5A = 1092

12A = 1092

Divide both sides by 12

A = 91

the weight of amberjack = 91 pounds

the weight of white marlin = 4A = 4(91) = 364 pounds

the weight of Tarpon = 2A = 2(91) = 182 pounds

the weight of Tuna= 5A = 5(91)=455 pounds

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Answer:

Segment EF: y = -x + 8

Segment BC: y = -x + 2

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Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.

Upon observing the given diagram, we can infer the following corresponding sides:

\displaystyle\mathsf{\overline{BC}\:\: and\:\:\overline{EF}}

\displaystyle\mathsf{\overline{BA}\:\: and\:\:\overline{ED}}

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We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.

<h2>Slope of Segment BC:</h2>

In order to solve for the slope of segment BC, we can use the following slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}  }

Use the following coordinates from the given diagram:

Point B:  (x₁, y₁) =  (-2, 4)

Point C:  (x₂, y₂) = ( 1,  1 )

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}

<h2>Slope of Segment EF:</h2>

Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:

Point E:  (x₁, y₁) =  (4, 4)

Point F:  (x₂, y₂) = (6, 2)

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}

Our calculations show that segment BC and EF have the same slope of -1.  In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.  

Since segments BC and EF have the same slope, then it means that  \displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}.

<h2>Slope-intercept form:</h2><h3><u>Segment BC:</u></h3>

The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.

Using the slope of segment BC, m = -1, and the coordinates of point C, (1,  1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>

y = mx + b

1 = -1( 1 ) + b

1 = -1 + b

Add 1 to both sides to isolate b:

1 + 1 = -1 + 1 + b

2 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.

Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:

⇒  y = -x + 2.

<h3><u /></h3><h3><u>Segment EF:</u></h3>

Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>

y = mx + b

4 = -1( 4 ) + b

4 = -4 + b

Add 4 to both sides to isolate b:

4 + 4 = -4 + 4 + b

8 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.

Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:

⇒  y = -x + 8.

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\mathrm{Group\:like\:terms}

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