Question

mcr3u unit 1 test due: september 24th @10:30am app / 18 t&i /23 1.calculate the perimeter. leave your answer in simplest form. 3a 2. determine the point(s) of intersection of the functions f(x)=4x² - x + 5 and g(x)=-41x - 91 algebraically. explain your answer. 5a

Explanation:

1.

Step1: Simplify $\sqrt{20}$

$\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$

Step2: Calculate the perimeter

The perimeter $P$ of the polygon is the sum of all its side - lengths. There are 4 sides of length $\sqrt{20}$ and 2 sides of length $3\sqrt{15}$.
$P = 4\sqrt{20}+2\times3\sqrt{15}$
Substitute $\sqrt{20}=2\sqrt{5}$ into the above formula:
$P = 4\times2\sqrt{5}+6\sqrt{15}=8\sqrt{5}+6\sqrt{15}$

Step1: Set $f(x)=g(x)$

Set $4x^{2}-x + 5=-41x-91$.
Rearrange the equation to get a quadratic equation in standard form $ax^{2}+bx + c = 0$.
$4x^{2}-x+41x + 5 + 91=0$, which simplifies to $4x^{2}+40x+96 = 0$.
Divide the entire equation by 4: $x^{2}+10x + 24=0$.

Step2: Solve the quadratic equation

Factor the quadratic equation $x^{2}+10x + 24=(x + 4)(x+6)=0$.
Set each factor equal to zero:
If $x + 4=0$, then $x=-4$.
If $x + 6=0$, then $x=-6$.

Step3: Find the corresponding $y$ - values

When $x=-4$, $y = g(-4)=-41\times(-4)-91=164 - 91 = 73$.
When $x=-6$, $y = g(-6)=-41\times(-6)-91=246-91 = 155$.
The points of intersection are $(-4,73)$ and $(-6,155)$.

Answer:

$8\sqrt{5}+6\sqrt{15}$

2.