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777dan777 [17]
3 years ago
11

How does 5x + 15 = 45 HELP PLZ

Mathematics
2 answers:
likoan [24]3 years ago
6 0

Answer:

When finding these problems we always have to do the opposite of the operation shown.

In this case its:

5x + 15 = 45

Subtract 15 from both sides (we start off with constants and numbers that are being shown) The reason why its subtraction is because since your adding 15 to 5x the opposite of that is subtraction so use the number shown by itself and subtract it from both sides.

5x = 30

Divide by 5 on both sides since the opposite of 5x is division (only use the number that is shown) We have to get rid of all the numbers on the left.

therefore you'll be left with:

X = 6

adelina 88 [10]3 years ago
3 0

Answer:

X=6

Step-by-step explanation:

5x+15=45

5x=45-15

5x=30,divide both sides by 5 to get x

x=6

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Which of the following represents the multiplicative inverse of the value below?
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Read 2 more answers
(a) Find the unit tangent and unit normal vectors T(t) and N(t).
andrey2020 [161]

Answer:

a. i. (i + tj + 2tk)/√(1 + 5t²)

ii.  (-5ti + j + 2k)/√[25t² + 5]

b. √5/[√(1 + 5t²)]³

Step-by-step explanation:

a. The unit tangent

The unit tangent T(t) = r'(t)/|r'(t)| where |r'(t)| = magnitude of r'(t)

r(t) = (t, t²/2, t²)

r'(t) = dr(t)/dt = d(t, t²/2, t²)/dt = (1, t, 2t)

|r'(t)| = √[1² + t² + (2t)²] = √[1² + t² + 4t²] = √(1 + 5t²)

So, T(t) = r'(t)/|r'(t)| = (1, t, 2t)/√(1 + 5t²)  = (i + tj + 2tk)/√(1 + 5t²)

ii. The unit normal

The unit normal N(t) = T'(t)/|T'(t)|

T'(t) = dT(t)/dt = d[ (i + tj + 2tk)/√(1 + 5t²)]/dt

= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + [-10tk/√(1 + 5t²)⁻³]

= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + j/√(1 + 5t²)+ [-10t²k/√(1 + 5t²)⁻³] + 2k/√(1 + 5t²)

= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) - 10t²k/[√(1 + 5t²)]⁻³ + 2k/√(1 + 5t²)

= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ - 10t²k/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) + 2k/√(1 + 5t²)

= -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + (j + 2k)/√(1 + 5t²)

We multiply by the L.C.M [√(1 + 5t²)]³  to simplify it further

= [√(1 + 5t²)]³ × -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + [√(1 + 5t²)]³ × (j + 2k)/√(1 + 5t²)

= -(i + tj + 2tk)5t + (j + 2k)(1 + 5t²)

= -5ti - 5²tj - 10t²k + j + 5t²j + 2k + 10t²k

= -5ti + j + 2k

So, the magnitude of T'(t) = |T'(t)| = √[(-5t)² + 1² + 2²] = √[25t² + 1 + 4] = √[25t² + 5]

So, the normal vector N(t) = T'(t)/|T'(t)| = (-5ti + j + 2k)/√[25t² + 5]

(b) Use Formula 9 to find the curvature.

The curvature κ = |r'(t) × r"(t)|/|r'(t)|³

since r'(t) = (1, t, 2t), r"(t) = dr'/dt = d(1, t, 2t)/dt  = (0, 1, 2)

r'(t) = i + tj + 2tk and r"(t) = j + 2k

r'(t) × r"(t) =  (i + tj + 2tk) × (j + 2k)

= i × j + i × 2k + tj × j + tj × 2k + 2tk × j + 2tk × k

= k - 2j + 0 + 2ti - 2ti + 0

= -2j + k

So magnitude r'(t) × r"(t) = |r'(t) × r"(t)| = √[(-2)² + 1²] = √(4 + 1) = √5

magnitude of r'(t) = |r'(t)| = √(1 + 5t²)

|r'(t)|³ = [√(1 + 5t²)]³

κ = |r'(t) × r"(t)|/|r'(t)|³ = √5/[√(1 + 5t²)]³

8 0
3 years ago
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