lavira1315
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I ONLY GOT ONE SHOT SO PLS HELP, IM STUPID! ANSWER IF YOU KNOW IT. OR AT LEAST TAKE A LOOK!!!!!!!! ( 2 QUESTIONS)


1. A rectangular trampoline measures 15 meters by 18 meters. A pad of constant width is placed around the trampoline so that the total area is 340 square meters.

What is the width of the pad?

MY GUESS IS 1 METER

2.
Use the quadratic formula.

2x^2−5x−9=0

Enter the solutions, in simplified radical form, in the boxes.

x =
or x =

Respuesta :

calculista

Answer:

Part 1) The width of the pad is [tex]1\ m[/tex]

Part 2) The solutions are

[tex]x=\frac{5(+)\sqrt{97}} {4}[/tex]  and [tex]x=\frac{5(-)\sqrt{97}} {4}[/tex]

Step-by-step explanation:

Part 1)

Let

x----> the width of the pad

we know that

[tex](15+2x)(18+2x)=340[/tex]

Solve for x

[tex]270+30x+36x+4x^{2} =340\\ \\4x^{2}+66x-70=0[/tex]

Solve the quadratic equation by graphing

The solution is [tex]x=1\ m[/tex]

see the attached figure

Part 2) we have

[tex]2x^{2} -5x-9=0[/tex]

we know that

The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to

[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]

in this problem we have

[tex]2x^{2} -5x-9=0[/tex]  

so

[tex]a=2\\b=-5\\c=-9[/tex]

substitute in the formula

[tex]x=\frac{5(+/-)\sqrt{-5^{2}-4(2)(-9)}} {2(2)}[/tex]

[tex]x=\frac{5(+/-)\sqrt{97}} {4}[/tex]

[tex]x=\frac{5(+)\sqrt{97}} {4}[/tex]

[tex]x=\frac{5(-)\sqrt{97}} {4}[/tex]

Ver imagen calculista

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