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WITCHER [35]
3 years ago
6

Solve for x in the equation X2-12x+36 = 90.

Mathematics
1 answer:
Setler79 [48]3 years ago
3 0

Answer:

90x8+3

Step-by-step explanation:

X2-12x+36 = 90.

X=

O x=6+3./10

O x=6+2/7

X= 123.22

O

x= 12+3/10

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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

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

6 0
3 years ago
Anwser this will give my prod
ddd [48]

Answer:

the answer is 9/2 = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2

Step-by-step explanation:

When yo add all the halves up you get 9/2

8 0
3 years ago
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Nastasia [14]

Answer:

L = 25 m

Step-by-step explanation:

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add 3 to each side:

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divide both sides by 5:

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6 0
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If m\angle A=(6x+10)^{\circ}∠A=(6x+10) ∘ and m\angle B=(4x+30)^{\circ}∠B=(4x+30) ∘ , then find the measure of \angle B∠B.​
netineya [11]

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Please help 30 points will give brainliest
Vinil7 [7]

Answer:

thats correct

Step-by-step explanation:

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