Question

1. solve for x and find m∠lmn. 4x + 2° 7x - 37° x = __ m∠lmn = __

Explanation:

Step1: Set up equation

Since the two - angle expressions form a right - angle (assuming the right - angle symbol in the figure), we know that $(4x + 2)+(7x-37)=90$.

Step2: Combine like terms

Combine the $x$ terms and the constant terms: $(4x+7x)+(2 - 37)=90$, which simplifies to $11x-35 = 90$.

Step3: Isolate the variable term

Add 35 to both sides of the equation: $11x-35 + 35=90 + 35$, resulting in $11x=125$.

Step4: Solve for x

Divide both sides by 11: $x=\frac{125}{11}\approx11.36$.

Step5: Find $m\angle LMN$

We can use either of the angle expressions. Let's use $4x + 2$. Substitute $x=\frac{125}{11}$ into $4x + 2$: $m\angle LMN=4\times\frac{125}{11}+2=\frac{500}{11}+2=\frac{500 + 22}{11}=\frac{522}{11}\approx47.45^{\circ}$ (using the first angle expression). Or using $7x-37$, $7\times\frac{125}{11}-37=\frac{875}{11}-37=\frac{875-407}{11}=\frac{468}{11}\approx42.55^{\circ}$. But since $(4x + 2)+(7x-37)=90$, if we take $4x + 2$:
$4x+2=4\times\frac{125}{11}+2=\frac{500 + 22}{11}=\frac{522}{11}\approx47.45^{\circ}$ and $7x-37=7\times\frac{125}{11}-37=\frac{875-407}{11}=\frac{468}{11}\approx42.55^{\circ}$. The sum is $47.45+42.55 = 90$.

Answer:

$x=\frac{125}{11}$; $m\angle LMN = 90^{\circ}$ (since the two angles together form a right - angle)