Question

1. geometry supplementary angles are two angles with measures that have a sum of 180°. complementary angles are two angles with measures that have a sum of 90°. the measure of the supplement of an angle is 10° more than twice the measure of the complement of the angle. let 90 - x equal the degree - measure of the complement angle and 180 - x equal the degree - measure of the supplement angle. write and solve an equation to find the measure of the angle.
2. geometry write and solve an equation to find the value of x so that the figures have the same area.
3. geometry write and solve an equation to find the value of x so that the figures have the same area.
4. geometry write and solve an equation to find the value of x so that the figures have the same area. the area of a trapezoid is \\(\frac{1}{2}h(b_1 + b_2)\\).

examples 6 and 7
solve each equation and state whether the equation has one solution, no solution, or is an identity.

1. - 6y - 3 = 3 - 6y
2. \\(\frac{1}{2}(x + 6)=\frac{1}{2}x - 9\\)
3. 8q + 12 = 4(3 + 2q)
4. 21(x + 1)-6x = 15x + 21
5. 12y + 48 - 4y = 8(y - 6)
6. 8(z + 6)=4(2z + 12)
7. 2a + 2 = 3(a + 2)
8. \\(\frac{1}{4}x+5 = \frac{1}{4}x\\)
9. 7(c + 9)=7c + 63
10. 4k + 3 = \frac{1}{4}(8k + 16)
11. 3b - 13 + 4b = 7b + 1
12. \\(\frac{1}{2}(\frac{1}{2}m - 8)=\frac{1}{4}(m - 16)\\)

Explanation:

21.

Step1: Set up the equation

Let the angle be $x$. The measure of its complement is $90 - x$ and the measure of its supplement is $180 - x$. Given that the measure of the supplement of an angle is 10° more than twice the measure of the complement. So, $180 - x=2(90 - x)+ 10$.

Step2: Expand the right - hand side

$180 - x = 180-2x + 10$.

Step3: Simplify the equation

Add $2x$ to both sides: $180 - x+2x=180 - 2x+2x + 10$, which gives $180 + x=180 + 10$.
Subtract 180 from both sides: $x = 10$.

Step1: Recall area formulas

The area of a triangle is $A_{triangle}=\frac{1}{2}\times base\times height=\frac{1}{2}\times8\times(x + 4)=4(x + 4)$. The area of a rectangle is $A_{rectangle}=length\times width=(x + 6)\times1$.

Step2: Set up the equation

Since the areas are equal, $4(x + 4)=x + 6$.

Step3: Expand the left - hand side

$4x+16=x + 6$.

Step4: Solve for $x$

Subtract $x$ from both sides: $4x - x+16=x - x + 6$, so $3x+16 = 6$.
Subtract 16 from both sides: $3x+16 - 16=6 - 16$, so $3x=-10$.
Divide both sides by 3: $x=-\frac{10}{3}$.

Step1: Recall area formulas

The area of the first rectangle is $A_1 = 9\times5=45$ square feet. The area of the second rectangle is $A_2=15.5\times x$.

Step2: Set up the equation

Since the areas are equal, $45 = 15.5x$.

Step3: Solve for $x$

Divide both sides by 15.5: $x=\frac{45}{15.5}=\frac{45\times2}{15.5\times2}=\frac{90}{31}\approx2.9$.

Answer:

$10^{\circ}$

22.