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(English Medium) ICSE Class 9 - CISCE Question Bank Solutions

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Evaluate :  `( log _5^8 )/(( log_25 16 ) xx  ( log_100  10))`

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

In a quadrilateral ABCD, AB = AD and CB = CD.

Prove that:

  1. AC bisects angle BAD.
  2. AC is the perpendicular bisector of BD.
[14] Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Chapter: [14] Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Concept: undefined >> undefined

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Prove that the bisectors of the interior angles of a rectangle form a square.

[14] Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Chapter: [14] Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Concept: undefined >> undefined

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ;

prove that AP and DQ are perpendicular to each other.

[14] Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Chapter: [14] Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Concept: undefined >> undefined

Evaluate: `(log_5 8)/(log_25 16 xx Log_100 10)` 

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log (3 - x) - log (x - 3) = 1

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log(x2 + 36) - 2log x = 1

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log 7 + log (3x - 2) =  log (x + 3) + 1

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log ( x + 1) + log ( x - 1) = log 11 + 2 log 3

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log 4 x + log 4 (x-6) = 2

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log 8 (x2 - 1) - log 8 (3x + 9) = 0

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
log (x + 1) + log (x - 1) = log 48

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve the following:
`log_2x + log_4x + log_16x = (21)/(4)`

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: log (x + 5) = 1

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: `("log"27)/("log"243)` = x

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: `("log"81)/("log"9)` = x

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: `("log"121)/("log"11)` = logx

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: `("log"125)/("log"5)` = logx

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: `("log"128)/("log"32)` = x

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined

Solve for x: `("log"1331)/("log"11)` = logx

[8] Logarithms
Chapter: [8] Logarithms
Concept: undefined >> undefined
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