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Anna [14]
2 years ago
7

Expand each expression A ( x+1)^2 B(x-3)^2. C ( x+4)^2 D (x-1/2)^2

Mathematics
1 answer:
Elis [28]2 years ago
3 0

The <em>expanded</em> expressions are shown below:

  1. (x + 1)² = x² + 2 · x + 1
  2. (x - 3)² = x² - 6 · x + 9
  3. (x + 4)² = x² + 8 · x + 16
  4. (x - 1/2)² = x² - x + 1/4
  5. 3 · (x - 5)² = 3 · (x² - 10 · x + 25) = 3 · x² - 30 · x + 75
  6. (1/2) · (x - 2)² = (1/2) · (x² - 4 · x + 4) = (1/2) · x² - 2 · x + 2

<h3>How to expand perfect square trinomials</h3>

<em>Perfect square</em> trinomials are polynomials of grade 2 with the form (a + b)² = a² + 2 · a · b + b². In this case, we have to expand six perfect square trinomials:

a) (x + 1)² = x² + 2 · x + 1

b) (x - 3)² = x² - 6 · x + 9

c) (x + 4)² = x² + 8 · x + 16

d) (x - 1/2)² = x² - x + 1/4

e) 3 · (x - 5)² = 3 · (x² - 10 · x + 25) = 3 · x² - 30 · x + 75

f) (1/2) · (x - 2)² = (1/2) · (x² - 4 · x + 4) = (1/2) · x² - 2 · x + 2

To learn more on perfect square trinomials: brainly.com/question/385286

#SPJ1

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x=15 cm

Step-by-step explanation:

The two triangles in the diagram are:

ABC and BDC

First we have to find the third side (hypotenuse) of BDC so that we can use it to find the value of x.

Hypotenuse is the largest side of a triangle which is usually in front of the right angle.

So in BDC

Base = BD =10cm\\Hypotenuse = BC = ?\\Perpendicular = CD = 2\sqrt{11}cm

Applying Pythagoras theorem:

(Hypotenuse)^2  = (Base)^2 + (Perpendicular)^2\\BC^2 = BD^2 + CD^2\\BC^2 = (10)^2 + (2\sqrt{11})^2\\BC^2 = 100+(2^2 * 11)\\BC^2 = 100+(4*11)\\BC^2 = 100+44\\BC^2 = 144\\\sqrt{BC^2} = \sqrt{144}\\BC = 12cm

Solving for triangle ABC

Base = BC = 12 cm\\Perpendicular = AB = 9 cm\\Hypotenuse = AC = x\\

Applying Pythagoras theorem

AC^2 = BC^2+AB^2\\x^2 = (12)^2 + (9)^2\\x^2 = 144+81\\x^2 = 225\\\sqrt{x^2} = \sqrt{225}\\x = 15cm

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Step-by-step explanation:

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