Question

30 points! i need to turn this in tomorrow

Answer 1

Explanation:

Reduce the radicals:

`1.\ √(72)=√(3*5*5)=√(3*5^2)\\\\\boxed{√(75)=5√(3) }`

`2.\ √(80)=√(2*2*2*2*5)=√(2^4*5)=2^2√(5)\\\\\boxed{√(80)=4√(5) }`

`3.\ √(108)=√(2*2*3*3*3)=√(2^2*3^2*3)=2*3√(3)\\\\\boxed{√(108)=6√(3)}`

`4.\ √(72)=√(2*2*2*3*3)=√(2^2*2*3^2)=2*3√(2)\\\\\boxed{√(72)=6√(72) }[`

Find the distance between the two points:

`5.\ (5, 9), (-7, -7)\\\\D=√((-7-5)^2+(-7-9)^2)=√((-12)^2+(-16)^2)=√(144+256)=√(400)=√(20^2)\\\\\boxed{D=20}`

`6.\ (0, -2), (-5, -1)\\\\D=√((-5-0)^2+(-1-(-2))^2)=√((-5)^2+(1)^2)=√(25+1)\\\\\boxed{D=√(26)}`

`7.\ D=10, A(4,w)\ and\ B(-2,-1)\\\\D=√((-2-4)^2+(-1-w)^2)\\\\10^2=[√((-6)^2+(-1-w)^2)]^2\\\\100=36+(-1-w)^2\\\\64=(-1-w)^2\\\\√(64) =√((-1-w)^2)\\\\\pm8=-1-w\\\\\pm8+1=-w\\\\w=\pm8-1\\\\\boxed{w_1=7\ and\ w_2=-9}`

`8.\ D=20, A(5,9)\ and\ B(-7,w);\ w<0\\\\D=√((-7-5)^2+(w-9)^2)\\\\20^2=[√((-12)^2+(w-9)^2)]^2\\\\400=144+(w-9)^2\\\\256=(w-9)^2\\\\√(256) =√((w-9)^2)\\\\\pm16=w-9\\\\\pm16+9=w\\\\\boxed{w=-7}`

Answer 2

9.
`√((4a)+(9b))` 10.
`√((9a)+(9b))` 11
`√(13)`

Explanation:

9. Since the Distance Formula is
`D=\sqrt{(x-x_(0))^(2)+(y-y_(0))^2}`

C(a, -b) D(3a, -4b)

Let us plug it in values

D=
`\sqrt{(3a-a)^(2)+(-4b+b)^(2)}\\\\\sqrt{(2a)^(2)+(-3b)^(2)}  \\ √((4a)+(9b))`

10. C(-a,-2b) D(2a, b)

D=
`\sqrt{(2a+a)^(2)+(b+2b)^(2)}\\\\\sqrt{(3a)^(2)+(3b)^(2)}  \\ √((9a)+(9b))`

11.

a)AB A(2,3) B(5,5)

D=
`\sqrt{(5-2)^(2)+(5-3)^(2)}\\\\\sqrt{(3)^(2)+(2)^(2)}  \\ √(13)`

CD C(4,3) D(1,1)

D=
`\sqrt{(1-4)^(2)+(1-3)^(2)}\\\\\sqrt{(-3)^(2)+(-2)^(2)}  \\ √(13)`

BC B(5,5) C(4,3)

D=
`\sqrt{(4-5)^(2)+(3-5)^(2)}\\\\\sqrt{(-1)^(2)+(-2)^(2)}  \\ √(5)`

DA D(1,1) A(2,3)

D=
`\sqrt{(2-1)^(2)+(3-1)^(2)}\\\\\sqrt{(1)^(2)+(2)^(2)}  \\ √(5)`

e. The length of each diagonal

AC A(2,3) C(4,3)

D=
`\sqrt{(4-2)^(2)+(3-3)^(2)}\\\\\sqrt{(2)^(2)+(0)^(2)}  \\ √(4)`

length = 2 u

f.BD B(5,5) D(1,1)

D=
`\sqrt{(1-5)^(2)+(1-5)^(2)}\\\\\sqrt{(-4)^(2)+(-4)^(2)}  \\ √(32)` = 4
`4√(2)`

g. No, since congruent diagonals have the same size. Those diagonals do not have congruence between them.

12.

AB A(-2,-1) B(4,1)

D=
`\sqrt{(4+2)^(2)+(1+1)^(2)}\\\\\sqrt{(6)^(2)+(2)^(2)}  \\ √(40)` =
`2√(10)`

AC A(-2,-1) C(2,-5)

D=
`\sqrt{(2+2)^(2)+(-5+1)^(2)}\\\\\sqrt{(4)^(2)+(-4)^(2)}  \\ √(32)` =
`4√(2)`

BC B(4,1) C(2,-5)

D=
`\sqrt{(2-4)^(2)+(-5-1)^(2)}\\\\\sqrt{(-2)^(2)+(-6)^(2)}  \\ √(40)` =
`2√(10)`

d) AB≅BC