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Sauron [17]
2 years ago
10

What is the surface area of the cube.A. 409 cm²B. 2400 cm²C. 2004 cm²D. 40 cm²​

Mathematics
1 answer:
larisa [96]2 years ago
4 0
  • Side=a=20cm

Surface area

  • 6a^2
  • 6(20)^2
  • 400(6)
  • 2400cm^2
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The measure of a central angle is 3x + 18 and the measure of its arc is 147 which equation can you use to find the value of x
steposvetlana [31]

3x + 18 = 147

3x = 129

x = 43

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There are 100 bolts and 2 wingnuts in each bookcase kit. What is the unit rate in bolts per wingnut?
gayaneshka [121]
Since they are asking for the unit rate in bolts per wing nut, you can think of it as bolts per ONE wing nut. 

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3 years ago
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Can somebody help me with my Geometry?​
bezimeni [28]

Answer:

∠KJL = 45°

∠x=45°

∠DBC=67.47°

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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

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(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
Need help on these asap please
Elenna [48]
9.6958 rounded to the nearest hundredth is 9.70
-
44cm
5 0
2 years ago
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