IDNLearn.com connects you with a community of knowledgeable individuals ready to help. Our platform provides detailed and accurate responses from experts, helping you navigate any topic with confidence.
Sagot :
Let's solve the given problem step-by-step:
The problem is to find the points at which the banner should be attached to the archway. The archway is modeled by the quadratic equation:
[tex]\[ y = -x^2 + 6x \][/tex]
The equation of the line representing the angle at which the banner is hung is:
[tex]\[ 4y = 21 - x \][/tex]
First, we need to express both equations in a form that can be easily solved simultaneously.
Starting with the second equation:
[tex]\[ 4y = 21 - x \][/tex]
Let's solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{21 - x}{4} \][/tex]
Now, we have two equations:
1. [tex]\( y = -x^2 + 6x \)[/tex]
2. [tex]\( y = \frac{21 - x}{4} \)[/tex]
We can set these expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -x^2 + 6x = \frac{21 - x}{4} \][/tex]
Next, let's eliminate the fraction by multiplying through by 4:
[tex]\[ 4(-x^2 + 6x) = 21 - x \][/tex]
[tex]\[ -4x^2 + 24x = 21 - x \][/tex]
Bring all terms to one side of the equation to form a standard quadratic equation:
[tex]\[ -4x^2 + 24x + x - 21 = 0 \][/tex]
[tex]\[ -4x^2 + 25x - 21 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -4 \)[/tex], [tex]\( b = 25 \)[/tex], and [tex]\( c = -21 \)[/tex].
Let's find the solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-25 \pm \sqrt{25^2 - 4(-4)(-21)}}{2(-4)} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{625 - 336}}{-8} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{289}}{-8} \][/tex]
[tex]\[ x = \frac{-25 \pm 17}{-8} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-25 + 17}{-8} = \frac{-8}{-8} = 1 \][/tex]
[tex]\[ x_2 = \frac{-25 - 17}{-8} = \frac{-42}{-8} = 5.25 \][/tex]
Now, substitute these values back into one of the original equations to find the corresponding [tex]\( y \)[/tex]-values:
Using [tex]\( y = -x^2 + 6x \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -(1)^2 + 6(1) = -1 + 6 = 5 \][/tex]
For [tex]\( x = 5.25 \)[/tex]:
[tex]\[ y = -(5.25)^2 + 6(5.25) = -(27.5625) + 31.5 = 3.9375 \][/tex]
Thus, the points where the banner should be attached to the archway are:
[tex]\[ (1, 5) \][/tex]
[tex]\[ (5.25, 3.94) \][/tex]
The correct answer is C:
C. [tex]\( (1, 5) \)[/tex] and [tex]\( (5.25, 3.94) \)[/tex]
The problem is to find the points at which the banner should be attached to the archway. The archway is modeled by the quadratic equation:
[tex]\[ y = -x^2 + 6x \][/tex]
The equation of the line representing the angle at which the banner is hung is:
[tex]\[ 4y = 21 - x \][/tex]
First, we need to express both equations in a form that can be easily solved simultaneously.
Starting with the second equation:
[tex]\[ 4y = 21 - x \][/tex]
Let's solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{21 - x}{4} \][/tex]
Now, we have two equations:
1. [tex]\( y = -x^2 + 6x \)[/tex]
2. [tex]\( y = \frac{21 - x}{4} \)[/tex]
We can set these expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[ -x^2 + 6x = \frac{21 - x}{4} \][/tex]
Next, let's eliminate the fraction by multiplying through by 4:
[tex]\[ 4(-x^2 + 6x) = 21 - x \][/tex]
[tex]\[ -4x^2 + 24x = 21 - x \][/tex]
Bring all terms to one side of the equation to form a standard quadratic equation:
[tex]\[ -4x^2 + 24x + x - 21 = 0 \][/tex]
[tex]\[ -4x^2 + 25x - 21 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -4 \)[/tex], [tex]\( b = 25 \)[/tex], and [tex]\( c = -21 \)[/tex].
Let's find the solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-25 \pm \sqrt{25^2 - 4(-4)(-21)}}{2(-4)} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{625 - 336}}{-8} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{289}}{-8} \][/tex]
[tex]\[ x = \frac{-25 \pm 17}{-8} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-25 + 17}{-8} = \frac{-8}{-8} = 1 \][/tex]
[tex]\[ x_2 = \frac{-25 - 17}{-8} = \frac{-42}{-8} = 5.25 \][/tex]
Now, substitute these values back into one of the original equations to find the corresponding [tex]\( y \)[/tex]-values:
Using [tex]\( y = -x^2 + 6x \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -(1)^2 + 6(1) = -1 + 6 = 5 \][/tex]
For [tex]\( x = 5.25 \)[/tex]:
[tex]\[ y = -(5.25)^2 + 6(5.25) = -(27.5625) + 31.5 = 3.9375 \][/tex]
Thus, the points where the banner should be attached to the archway are:
[tex]\[ (1, 5) \][/tex]
[tex]\[ (5.25, 3.94) \][/tex]
The correct answer is C:
C. [tex]\( (1, 5) \)[/tex] and [tex]\( (5.25, 3.94) \)[/tex]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.
