Question

30 points! i need to turn this in tomorrow

Answer 1

Answer:

Step-by-step explanation:

Reduce the radicals:

$1.\ \sqrt{72}=\sqrt{3\times5\times5}=\sqrt{3\times5^2}\\\\\boxed{\sqrt{75}=5\sqrt{3} }$

$2.\ \sqrt{80}=\sqrt{2\times2\times2\times2\times5}=\sqrt{2^4\times5}=2^2\sqrt{5}\\\\\boxed{\sqrt{80}=4\sqrt{5} }$

$3.\ \sqrt{108}=\sqrt{2\times2\times3\times3\times3}=\sqrt{2^2\times3^2\times3}=2\times3\sqrt{3}\\\\\boxed{\sqrt{108}=6\sqrt{3}}$

$4.\ \sqrt{72}=\sqrt{2\times2\times2\times3\times3}=\sqrt{2^2\times2\times3^2}=2\times3\sqrt{2}\\\\\boxed{\sqrt{72}=6\sqrt{72} }[$

 

Find the distance between the two points:

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

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

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

 

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

Answer 2

Answer:

9. $\sqrt{(4a)+(9b)}$ 10. $\sqrt{(9a)+(9b)}$ 11$\sqrt{13}$

Step-by-step 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=$\sqrt{(3a-a)^{2}+(-4b+b)^{2}}\\\\\sqrt{(2a)^{2}+(-3b)^{2}}  \\ \sqrt{(4a)+(9b)}$

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

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

11.

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

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

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

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

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

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

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

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

e. The length of each diagonal

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

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

length = 2 u

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

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

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=$\sqrt{(4+2)^{2}+(1+1)^{2}}\\\\\sqrt{(6)^{2}+(2)^{2}}  \\ \sqrt{40}$ = $2\sqrt{10}$

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

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

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

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

d) AB≅BC