Please help me

Mathematics

2 answers:

Lyrx [107]2 years ago

5 0

Step-by-step explanation:

d^2=(2^2)+(5^2)

=4+25

=29

d=√29 =4.58units

Paraphin [41]2 years ago

4 0

Answer:

♛┈⛧┈┈•༶♛┈⛧┈┈•༶

You might be interested in

A $2,700 principal earns 5% interest, compounded annually. After 4 years, what is the balance in the account

natulia [17]

Answer:

Step-by-step explanation:

67001

7 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

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