Answer :
To determine the first day that at least 100 masks were sold during the fundraiser, we start with the given equation modeling the number of masks sold each day:
[tex]\[ m = d^2 - 12d + 45 \][/tex]
We need to find the value of [tex]\( d \)[/tex] such that the number of masks sold [tex]\( m \)[/tex] is at least 100. In other words, we need to solve the inequality:
[tex]\[ d^2 - 12d + 45 \geq 100 \][/tex]
First, we subtract 100 from both sides of the inequality to set it to zero:
[tex]\[ d^2 - 12d + 45 - 100 \geq 0 \][/tex]
[tex]\[ d^2 - 12d - 55 \geq 0 \][/tex]
Next, we solve the corresponding equation:
[tex]\[ d^2 - 12d - 55 = 0 \][/tex]
To find the solutions to this quadratic equation, we can use the quadratic formula:
[tex]\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this case, the coefficients are [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -55 \)[/tex]. Substituting these values into the quadratic formula, we get:
[tex]\[ d = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-55)}}{2 \cdot 1} \][/tex]
[tex]\[ d = \frac{12 \pm \sqrt{144 + 220}}{2} \][/tex]
[tex]\[ d = \frac{12 \pm \sqrt{364}}{2} \][/tex]
[tex]\[ d = \frac{12 \pm \sqrt{4 \cdot 91}}{2} \][/tex]
[tex]\[ d = \frac{12 \pm 2\sqrt{91}}{2} \][/tex]
[tex]\[ d = 6 \pm \sqrt{91} \][/tex]
The two solutions for [tex]\( d \)[/tex] are:
[tex]\[ d = 6 + \sqrt{91} \][/tex]
[tex]\[ d = 6 - \sqrt{91} \][/tex]
Since [tex]\( d \)[/tex] represents a day in the fundraiser, which must be a positive integer, we discard the negative solution [tex]\( d = 6 - \sqrt{91} \)[/tex] (as [tex]\( \sqrt{91} \)[/tex] is approximately 9.5, making this solution negative).
The smallest positive solution is:
[tex]\[ d = 6 + \sqrt{91} \][/tex]
Given that [tex]\( \sqrt{91} \)[/tex] is approximately 9.5, the actual value [tex]\( 6 + \sqrt{91} \)[/tex] is in the domain of possible days in the interval of the first three weeks. For practical purposes, taking the first positive integer day that fits [tex]\( d \geq 6 + \sqrt{91} \)[/tex], we identify the answer fitting the problem's context.
Hence, 100 masks were sold for the first time on the [tex]\( \boxed{6+\sqrt{91}} \)[/tex] th day.
[tex]\[ m = d^2 - 12d + 45 \][/tex]
We need to find the value of [tex]\( d \)[/tex] such that the number of masks sold [tex]\( m \)[/tex] is at least 100. In other words, we need to solve the inequality:
[tex]\[ d^2 - 12d + 45 \geq 100 \][/tex]
First, we subtract 100 from both sides of the inequality to set it to zero:
[tex]\[ d^2 - 12d + 45 - 100 \geq 0 \][/tex]
[tex]\[ d^2 - 12d - 55 \geq 0 \][/tex]
Next, we solve the corresponding equation:
[tex]\[ d^2 - 12d - 55 = 0 \][/tex]
To find the solutions to this quadratic equation, we can use the quadratic formula:
[tex]\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this case, the coefficients are [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -55 \)[/tex]. Substituting these values into the quadratic formula, we get:
[tex]\[ d = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-55)}}{2 \cdot 1} \][/tex]
[tex]\[ d = \frac{12 \pm \sqrt{144 + 220}}{2} \][/tex]
[tex]\[ d = \frac{12 \pm \sqrt{364}}{2} \][/tex]
[tex]\[ d = \frac{12 \pm \sqrt{4 \cdot 91}}{2} \][/tex]
[tex]\[ d = \frac{12 \pm 2\sqrt{91}}{2} \][/tex]
[tex]\[ d = 6 \pm \sqrt{91} \][/tex]
The two solutions for [tex]\( d \)[/tex] are:
[tex]\[ d = 6 + \sqrt{91} \][/tex]
[tex]\[ d = 6 - \sqrt{91} \][/tex]
Since [tex]\( d \)[/tex] represents a day in the fundraiser, which must be a positive integer, we discard the negative solution [tex]\( d = 6 - \sqrt{91} \)[/tex] (as [tex]\( \sqrt{91} \)[/tex] is approximately 9.5, making this solution negative).
The smallest positive solution is:
[tex]\[ d = 6 + \sqrt{91} \][/tex]
Given that [tex]\( \sqrt{91} \)[/tex] is approximately 9.5, the actual value [tex]\( 6 + \sqrt{91} \)[/tex] is in the domain of possible days in the interval of the first three weeks. For practical purposes, taking the first positive integer day that fits [tex]\( d \geq 6 + \sqrt{91} \)[/tex], we identify the answer fitting the problem's context.
Hence, 100 masks were sold for the first time on the [tex]\( \boxed{6+\sqrt{91}} \)[/tex] th day.
