PLEAZZZZZZZZZZ HELP ASAP One base of the trapizoid is three times the lengh of the second base. The height of the trapizoid is 2 in. smaller than the second base. If the area of the trapizoid is 30in^2, find the lengths of the bases and the height of the trapizoid . A student is painting an accent wall in his room where the length of the wall is 3ft^2 more than its width. The wall in his room where the length is 3ft more than its width . The wall has an area of 130ft^2. What are the length and the width, in feet.Find two consecutive even integers whose product is 80. (there are two pairs, and only an algebraic solution will be excepted)

QUESTION 1

GIVEN:

[tex]long \: base = 3 \times (short \: base)[/tex]

[tex]height = (short \: base) - 2[/tex]

[tex]area = 30 {in}^{2} [/tex]

IMPORTANT EQUATIONS

[tex]trapezoid \: area = \frac{sum \: of \: bases}{2} \times height[/tex]

SOLVE:

[tex]30 = \frac{short + long}{2} (short - 2)[/tex]
[tex]30 = \frac{(short + 3short)}{2} \times (short - 2)[/tex]

[tex]30 = \frac{4short}{2} (short - 2)[/tex]

[tex]30 = 2{(short)}^{2} - 4(short)[/tex]

[tex] {short}^{2} - 2(short) - 15 = 0[/tex]

factor:

[tex](short - 5)(short + 3) = 0[/tex]

[tex]short = 3 \: and \: - 5[/tex]
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

[tex]short \: base = 3 \: in[/tex]

[tex]long \: base = 3 \times 3 = 9 \: in[/tex]

[tex]height = 3 - 2 = 1 \: in[/tex]

QUESTION 2

GIVEN
convert 3ft to inches.

[tex]l = 32 + w[/tex]

[tex]area = lw = 130 {in}^{2} [/tex]
or

[tex]w = \frac{130}{l} [/tex]

substitute

[tex]l = 32 + \frac{130}{l} [/tex]

solve for l:

[tex] {l}^{2} - 32 l- 130 = 0[/tex]
QUESTION 3

[tex]x(x + 2) = 80[/tex]

solve for X

[tex] {x}^{2} + 2x - 80 = 0[/tex]
[tex](x - 10)(x + 8) = 0[/tex]

[tex]x = 10 \: and \: - 8[/tex]

take the positive value since it's the only one that makes sense in this context.

so the two values are:

[tex]10 \: and \: 8[/tex]