Question

1. (5 - 29) review: pre - algebra

solve the following equations for the given variable, if possible. remember to check your answers.
a. 6x² = 150
b. 4m + 3 - m = 3(m + 1)

1. (5 - 30) use two different trig ratios to find the measure of ∠a. did you get the same answer both ways?

(5 - 33) find the area and the perimeter of the figure at right. be sure to organize your work so you can explain your method later.

Explanation:

1a. Solve the equation $6x^{2}=150$ for $x$

Step1: Isolate $x^{2}$

Divide both sides of the equation by 6.
$x^{2}=\frac{150}{6}=25$

Step2: Solve for $x$

Take the square - root of both sides. Remember, when taking the square root, we have two solutions: positive and negative.
$x=\pm\sqrt{25}=\pm5$

Step1: Simplify the left - hand side

Combine like terms on the left - hand side. $4m - m+3=3m + 3$.
The left - hand side is $3m + 3$.

Step2: Simplify the right - hand side

Distribute the 3 on the right - hand side: $3(m + 1)=3m+3$.

Step3: Analyze the equation

We have the equation $3m + 3=3m + 3$. This is an identity, which means it is true for all real values of $m$.

Step1: Use the sine ratio

The sine of an angle in a right - triangle is defined as $\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle A$, the opposite side is 12 and the hypotenuse is 13. So, $\sin(A)=\frac{12}{13}$. Then $A=\sin^{-1}(\frac{12}{13})\approx67.38^{\circ}$

Step2: Use the cosine ratio

The cosine of an angle in a right - triangle is defined as $\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle A$, the adjacent side is 5 and the hypotenuse is 13. So, $\cos(A)=\frac{5}{13}$. Then $A=\cos^{-1}(\frac{5}{13})\approx67.38^{\circ}$

Answer:

$x = 5$ or $x=-5$

1b. Solve the equation $4m + 3 - m=3(m + 1)$ for $m$