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Step-by-step explanation:
5.
( 1 ) A + B
⇒ [ 5x² - 3y² + z² ] + [ -2x² + 3y² -4z² ]
⇒ 3x² - 3z²
( 2 ) A - B
⇒ [ 5x² - 3y² + z² ] - [ -2x² + 3y² -4z² ]
⇒ 7x² - 6y² + 5z²
6. Given,
POQ is a straight line
∴ ∠ POQ = 180°
∠POQ = ∠POR + ∠QOR
180° = 3x + 2x + 10
170° = 5x
x = 170° / 5
x = 34°
∴ x = 34°
7. Given,
height = base/2
Area = 144 cm²
Area = 1/2 × b × h
144 cm² = 1/2 × b × h
144 cm² / 2 = b × b / 2
144 cm² × 2 / 2 = b²
144 cm² = b²
b = √144 cm²
b = 12 cm
∴ Length of height = b / 2 = 12 / 2 = 6 cm & Length of base = 12 cm
8.
( 1 ) a^b + b^-a
⇒ ( -1 )² + ( 2 )^-1
⇒ 1 + 1/2
⇒ 3 / 2
( 2 ) a^-a + b^b
⇒ ( - 1 )^-1 + ( 2^2 )
⇒ 1 / - 1 + 4
⇒ -1 + 4
⇒ 3
EXPLANATION.
Question = 1.
A = 5x² - 3y² + z² and B = -2x² + 3y² - 4z².
(a) = A + B.
⇒ 5x² - 3y² + z² + (-2x² + 3y² - 4z²).
⇒ 5x² - 3y² + z² - 2x² + 3y² - 4z².
⇒ 5x² - 2x² - 3y² + 3y² + z² - 4z².
⇒ 3x² - 3z².
⇒ 3(x² - z²).
⇒ 3(x - z)(x + z).
(b) = A - B.
5x² - 3y² + z² - (-2x² + 3y² - 4z²).
5x² - 3y² + z² + 2x² - 3y² + 4z².
5x² + 2x² - 3y² - 3y² + z² + 4z².
7x² - 6y² + 5z².
Question = 2.
∠POR = 3x and ∠QOR = 2x + 10.
POQ is a straight lines.
As we know that,
⇒ ∠POR + ∠QOR = 180°.
⇒ 3x + 2x + 10 = 180°.
⇒ 5x + 10 = 180°.
⇒ 5x = 180 - 10.
⇒ 5x = 170°.
⇒ x = 34°.
Question = 3.
In a triangle (Δ),
The height is half of the base.
Area = 144cm².
As we know that,
Area of triangle (Δ) = 1/2 x height x base.
⇒ h = b/2.
⇒ 144 = 1/2 x b/2 x b.
⇒ 144 = b²/4.
⇒ 576 = b².
⇒ b = 24cm.
⇒ h = b/2 = 24/2 = 12cm.
Question = 4.
⇒ a = -1 and b = 2.
(1) = a^(b) + b^(-a).
Put the values in the equation, we get.
⇒ (-1)² + (2)⁽⁻¹⁾.
⇒ 1 + 1/2.
⇒ 2 + 1/2 = 3/2.
(2) = a^(-a) + b^(b).
Put the values in the equation, we get.
⇒ (1)^(-1) + (2)^(2).
⇒ 1 + 4 = 5.