Mathematics, 29.11.2020 18:30 1060975

Hi . Please I need help with these questions :
See image for question.
Answer no 5 and 6.

Answers: 2

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Answers

Answer from: Redhead667

(i) √23

(ii) 196√23

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Standard Form: ax² + bx + c = 0Quadratic Formula: $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Step-by-step explanation:

Step 1: Define

Standard Form: x² - 10x + 2 = 0

Step 2: Define variables

a = 1

b = -10

c = 2

Step 3: Find roots

Substitute: $x=\frac{10\pm\sqrt{(-10)^2-4(1)(2)} }{2(1)}$ Exponents: $x=\frac{10\pm\sqrt{100-4(1)(2)} }{2(1)}$ Multiply: $x=\frac{10\pm\sqrt{100-8} }{2}$ Subtract: $x=\frac{10\pm\sqrt{92} }{2}$ Simplify: $x=\frac{10\pm 2\sqrt{23} }{2}$ Factor: $x=\frac{2(5\pm \sqrt{23}) }{2}$ Divide: $x=5\pm \sqrt{23}$

Step 4: Define roots

α > β

α = 5 + √23

β = 5 - √23

Step 5: Evaluate

Substitute: $\frac{1}{5-\sqrt{23} } -\frac{1}{5+\sqrt{23} }$ Subtract: $\sqrt{23}$

Substitute: $(5+\sqrt{23})^3-(5-\sqrt{23})^3$ Evaluate: $(98\sqrt{23} +470)-(470 - 98\sqrt{23} )$ Subtract: $196\sqrt{23}$

Answer from: elijahcarson9015

(i) √23(i) 196√23

Step-by-step explanation:

Given equation:

x² - 10x + 2 = 0

Roots are:

α and β

Sum of the roots:

α + β = -b/a ⇒ α + β = -(-10)/1 ⇒ α + β = 10

Product of the roots:

αβ = c/a ⇒ αβ = 2/1 ⇒ αβ = 2

Finding the following:

(i)

1/β - 1/α =(α - β)/αβ =√(α - β)² / αβ =√((α + β)² - 4αβ) / αβ = √(10² - 4*2) / 2 =√92 / 2 =2√23 / 2 =√23

(ii)

α³ - β³ =(α - β)(α² + αβ + β²) =√(α - β)²× ((α + β)² - αβ)√((α + β)² - 4αβ) × ((α + β)² - αβ) =√(10² - 4*2) × (10² -2) =√92 × 98 =2√23 × 98 =196√23

Answer from: Quest

i could probably can you me with a question of mine?

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

= $\sqrt{(9-6)^{2} +(12-6)^{2} } \\=\sqrt(45)=3\sqrt(5)$

side j'k'=distance between j' and k'

= $\sqrt{(3-2)^{2} +(2-4)^{2} } \\=sqrt(5)$

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

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Hi . Please I need help with these questions :
See image for question.
Answer no 5 and 6...