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frez [133]
3 years ago
6

kendall knows that a 45-ounce pitcher can hold enough lemonade for 6 people. At this rate, how many ounces of lemonade will kend

all need to serve to 26 people? HELP?
Mathematics
2 answers:
Angelina_Jolie [31]3 years ago
5 0
THIS IS A PROPORTION QUESTION. 

45       x                   CROSS MULTIPLY.
__  = ____
6        26

45*26=1170
6*x=6x


1170=6x                    DIVIDE 6 ON BOTH SIDES
_________
195=x


THE OFFICIAL ANSWER IS 195 OUNCES OF LEMONADE WOULD BE NEEDED TO SERVE 26 PEOPLE. HOPE THIS HELPS!!!!!!!!!! :)
nydimaria [60]3 years ago
5 0

Answer:195 ounce

Step-by-step explanation:

45 ounce = 6people

x =26people

cross multiply

6x = 45×26

6x=1170

Divide bothside by 6

6x/6 = 1179/6

x= 195

Therefore 195 ounce of lemonade is needed to serve 26 people

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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}}

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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:

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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}

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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

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1 + 1 = -1 + 1 + b

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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:

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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>

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

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