Use the given information to prove that RSU=TSV

Mathematics

1 answer:

RoseWind [281]2 years ago

3 0

Answer:

Statement, Reason

Given segment RV = segment TU, Given

m∠R = m∠T, Given

Segment RS ≅ segment ST, Base angles theorem

ΔRSV ≅ ΔTSU, SAS rule of congruency

Segment SV ≅ segment SU, CPCTC

VU ≅ UV, Reflexive property

RU = RV + VU, TV = TU + UV, Addition of segments

RV + VU ≅ TU + UV, Addition property of equality

RU ≅ TV, Transitive property of addition

ΔRSU ≅ ΔTSV, SSS rule of congruency

Step-by-step explanation:

Where:

SAS- Side Angle Side

CPCTC -Congruent Parts of Congruent Triangles are Congruent

SSS -Side Side Side

You might be interested in

Mr.smith has a maximum of $50 to spend at a museum a ticket to the museum cost $7 he can spend p dollars to buy other things at

aliya0001 [1]

Answer:

brainly.com/question/11804436

Step-by-step explanation:

USE LINK!!!!

5 0

2 years ago

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2

Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

$y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\$

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t = e^−2(0) = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

$\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt} = -2te^{-2t} = -2(0)e^{-2(0)} = 0$