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Taya2010 [7]
3 years ago
10

The function f(x) = –x2 + 20x – 75 models the profit from one customer, in dollars, a shop makes for printing photos, where x is

the number of photos printed, and f(x) is the amount of profit.
Part A: Determine the vertex. What does this calculation mean in the context of the problem? (5 points)

Part B: Determine the x-intercepts. What do these values mean in the context of the problem? (5 points)

Mathematics
1 answer:
trasher [3.6K]3 years ago
8 0
The function is graphed as shown below

Part A:

We use the formula - \frac{b}{2a} to find the vertex of the function. A quadratic function of the form of a x^{2} +bx+c and equating this form to the given function - x^{2} +20x-75, we have
b=20 and a=-1.

Substituting a and b into the vertex formula, we have
x=- \frac{20}{2(-1)}=10, as shown in the graph

This calculation means that the highest profit is achieved when the number of photo printed equals to ten photos

Part B:

We can find solution to this equation by factorising

- x^{2} +20x-75=0
-( x^{2} -20x+75)=0
x^{2} -20x+75=0
(x-5)(x-15)=0
x=5 and x=15, as shown in the graph

The two values means that the company makes no profit when they either produce 5 or 15 photos

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LOTS OF POINTS GIVING BRAINLIEST I NEED HELP PLEASEE
Sidana [21]

Answer:

Segment EF: y = -x + 8

Segment BC: y = -x + 2

Step-by-step explanation:

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}}

\displaystyle\mathsf{\overline{AC}\:\: and\:\:\overline{DF}}

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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<em></em>

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We know that the sum of three angles in a triangle is equal to 180^\circ.

\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 27+90+\angle C=180^\circ\\\Rightarrow \angle C = 63^\circ

The dimensions given in \triangle WXY are:

\angle Y = 63^\circ\\\angle X = 90^\circ

We know that the sum of three angles in a triangle is equal to 180^\circ.

\angle W+\angle X+\angle Y = 180^\circ\\\Rightarrow \angle W+90+63=180^\circ\\\Rightarrow \angle W = 27^\circ

Now, if we compare the angles of the two triangles:

\angle A = \angle W = 27^\circ\\\angle B = \angle X= 90^\circ\\\angle C = \angle Y= 63^\circ

So, by AA postulate (i.e. Angle - Angle) postulate, the two triangles are similar.

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Answer:

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