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Natalka [10]
3 years ago
13

What is the solution to 5x < 30

Mathematics
1 answer:
max2010maxim [7]3 years ago
7 0

Let's solve your inequality step-by-step.

5x<30

Step 1: Divide both sides by 5.

5x

5

<

30

5

x<6

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Use synthetic division to solve (x^3+1)/(x-1). What is the quotient?
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  x^2 +x +1

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The remainder is 2.

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g 1.32 Two points on a sphere of radius 3 are given as P1(3,0,30) and P2(3,45,45): (a) Find the position vectors of P1 and P2. (
Rama09 [41]

Answer:

a) P.V  of is OP₁ = [ 1.5i + 0j + 2.6k ],   P.V  of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b) Vector connecting P₁ to P₂ is [ 0i + 1.5j + 0.48k ]  

c) cylindrical coordinates are (1.5, π/2, 0.48)

Step-by-step explanation:

Given that;

r = 3

P₁ ( 3, 0°, 30° ),   P₂ ( 3, 45°, 45° )

a)

P.V of P₁

x = rcos∅sin∅ = 3(cos0°) ( sin30°) = (3 × 1 × 0.5) = 1.5

y = rsin∅sin∅  = 3(sin0°) (sin30°)   = (3 × 0 × 0.5) = 0

z = rcos∅        = 3(cos30°)             = ( 3 × 0.866)  = 2.6

∴ P.V  of is OP₁ = [ 1.5i + 0j + 2.6k ]

P.V of P₂

x = rcos∅sin∅ = 3(cos45°) ( sin45°) = (3 × 0.7071 × 0.7071) = 1.5

y = rsin∅sin∅  = 3(sin45°) (sin45°)   = (3 × 0.7071 × 0.7071) = 1.5

z = rcos∅        = 3(cos45°)                 = ( 3 × 0.7071)            = 2.12

∴ P.V  of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b)

Vector connecting P₁ to P₂ is given by

OP₂ - OP₁ = [ 1.5i + 1.5j + 2.12k ] - [ 1.5i + 0j + 2.6k ]

= [ 0i + 1.5j + 0.48k ]  

c)

P₁P₂ → = [ 0i + 1.5j + 0.48k ]  = [ 1.5j + 0.48k ]  

so in a cylindrical coordinate, it should be

r = √(o² + 1.5²) = 1.5

∅ = tan⁻¹[y/π] = π/2

z = 0.48

cylindrical coordinates are (1.5, π/2, 0.48)

5 0
3 years ago
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