What is the answer ??
UD = ?

Answer:

UD = 20

Step-by-step explanation:

Given a tangent and a secant from an external point to the circle, then

The square of the tangent is equal to the product of the external part and the entire secant, that is

12² = 7.2(7.2 + CD)

144 = 7.2² + 7.2CD

144 = 51.84 + 7.2CD ( subtract 51.84 from both sides )

92.16 = 7.2CD ( divide both sides by CD )

12.8 = CD

Thus

UD = UC + CD = 7.2 + 12.8 = 20

Need help with number 7 plzzz and just zoom in if u can’t see and I’ll love u forever

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Algebra 2 - Grade 11


Can anyone please help me with these Algebra 2 Math questions?


Soon as possible !!!



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Evaluate when x = -3 and y= -1:

NEED HELP NOW! 5 questions, 76 points!


1.
Given points B(−12,2), T(−8,10), A(−7,12), F(−6,10), and G(−2,2), which of the following proves that △BAG~△TAF?


2.
Given points A(2,1), B(1,4), C(−1,10), D(8,13), and E(4,5), which of the following proves that △ABE~△ACD?
Answers:

By the Distance Formula,AB=310‾‾‾√,AC=10‾‾‾√, AE=65‾√, and AD=25‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SAS ∼.


By the Distance Formula,AB=310‾‾‾√,AC=10‾‾‾√, AE=65‾√, and AD=25‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SSS ∼.

By the Distance Formula,AB=10‾‾‾√,AC=310‾‾‾√, AE=25‾√, and AD=65‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SSS ∼.

By the Distance Formula,AB=10‾‾‾√,AC=310‾‾‾√, AE=25‾√, and AD=65‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SAS ∼.



3.
Given points M(1,2), N(4,4), P(5,−1), J(−7,−1), K(−1,3), and L(1,−7), which of the following proves that △MNP≈△JKL?
Answers:

By the Distance Formula,MN=13‾‾‾√,NP=26‾‾‾√, and PM=5.Also, JK=213‾‾‾√,KL=226‾‾‾√, and LJ=10.Therefore, MNJK=NPKL=PMLJ=12, andtherefore, △MNP∼△JKL by SAS ∼.

By the Distance Formula,MN=13‾‾‾√,NP=26‾‾‾√, and PM=5.Also, JK=213‾‾‾√,KL=226‾‾‾√, and LJ=10.Therefore, MNJK=NPKL=PMLJ=12, andtherefore, △MNP∼△JKL by SSS ∼.

By the Distance Formula,MN=13,NP=26, and PM=25.Also, JK=52,KL=104, and LJ=100.Therefore, MNJK=NPKL=PMLJ=14, andtherefore, △MNP∼△JKL by SSS ∼.

By the Distance Formula,MN=13,NP=26, and PM=25.Also, JK=52,KL=104, and LJ=100.Therefore, MNJK=NPKL=PMLJ=14, andtherefore, △MNP∼△JKL by SAS ∼.



4.
Given points M(−5,5), R(−2,6), I(−2,4), B(2,−2), A(8,0), and G(8,−4), which of the following proves that △MRI~△BAG?
Answers:

By the Distance Formula,MR=10‾‾‾√,RI=2, and IM=10‾‾‾√.Also, BA=210‾‾‾√,AG=4, and GB=210‾‾‾√.Therefore, MRBA=RIAG=IMGB=10√210√=12,and therefore, △MRI~△BAG by SSS ~.

By the Distance Formula,MR=10,RI=2, and IM=10.Also, BA=40,AG=8, and GB=40.Therefore, MRBA=RIAG=IMGB=1040=14,and therefore, △MRI∼△BAG by SSS ∼.

By the Distance Formula,MR=2,RI=10, and IM=10.Also, BA=8,AG=40, and GB=40.Therefore, MRBA=RIAG=IMGB=1040=14,and therefore, △MRI∼△BAG by SAS ∼.

By the Distance Formula,MR=10‾‾‾√,RI=2, and IM=10‾‾‾√.Also, BA=210‾‾‾√,AG=4, and GB=210‾‾‾√.Therefore, MRBA=RIAG=IMGB=10√210√=12,and therefore, △MRI∼△BAG by SAS ∼.



5.
Given points L(−2,−2), A(−6,−3), U(−14,−5), X(4,−5), and P(0,−3), which of the following proves that △XUL~△PAL?

By the Distance Formula,LX=317‾‾‾√,PL=17‾‾‾√, LA=5‾√, and LU=35‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SSS ∼.

By the Distance Formula,LX=35‾√,PL=5‾√, LA=17‾‾‾√, and LU=317‾‾‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SAS ∼.

By the Distance Formula,LX=35‾√,PL=5‾√, LA=17‾‾‾√, and LU=317‾‾‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SSS ∼.

By the Distance Formula,LX=317‾‾‾√,PL=17‾‾‾√, LA=5‾√, and LU=35‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SAS ∼.

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