For how many positive values of n are both n3 and 3n four-digit integers?
1 answer:
Answer:
No positive value of n
Step-by-step explanation:
we have to find out for how many positive values of n are both
our-digit integers
Let us consider first cube
we get 4digit lowest number is 1000 and it has cube root as 10.
Thus 10 is the least integer which satisfies four digits for cube.
The highest integer is 9999 and it has cube root as 21.54
or 21 the highest integer.
Considering 3^n we get,
3^10 is having 5 digits and also 3^21
Thus there is no positive value of n which satisfy that both n cube and 3 power n are four digits.
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The missing figure is attached
(i) m∠APB = 90° ⇒ proved
(ii) AD = DP and PB = PC = BC ⇒ proved
(iii) DC = 2 AD ⇒ proved
Step-by-step explanation:
ABCD is a parallelogram in which:
- m∠A = 60°
- The bisectors of ∠A and ∠B meet DC at P
In parallelogram ABCD:
∵ m∠A = m∠C ⇒ opposite angles
∵ m∠A = 60°
∴ m∠C = 60°
∵ m∠A + m∠B = 180 ⇒ two adjacent supplementary angles
∴ 60 + m∠B = 180 ⇒ subtract 60 from both sides
∴ m∠B = 120°
∵ m∠B = m∠D ⇒ opposite angles
∴ m∠D = 120°
∵ AP is the bisector of angle A
- That means AP divide ∠A into two equal parts
∴ m∠BAP = m∠DAP = m∠A
∴ m∠BAP = m∠DAP = (60°)
∴ m∠BAP = m∠DAP = 30°
∵ BP is the bisector of angle B
- That means BP divide ∠B into two equal parts
∴ m∠ABP = m∠CBP = m∠B
∴ m∠ABP = m∠CBP = (120°)
∴ m∠ABP = m∠CBP = 60°
(i)
In ΔAPB
∵ m∠BAP = 30° ⇒ proved
∵ m∠ABP = 60° ⇒ proved
- Sum of the measures of the interior angles in a Δ is 180°
∴ m∠APB + m∠BAP + m∠ABP = 180°
∴ m∠APB + 30 + 60 = 180
- Add like terms in the left hand side
∴ m∠APB + 90 = 180
- Subtract 90 from both sides
∴ m∠APB = 90°
(ii)
In Δ ADP:
∵ m∠D = 120° ⇒ Proved
∵ m∠DAP = 30° ⇒ proved
- Sum of the measures of the interior angles in a Δ is 180°
∴ m∠APD + m∠DAP + m∠D = 180°
∴ m∠APD + 30 + 120 = 180
- Add like terms in the left hand side
∴ m∠APD + 150 = 180
- Subtract 150 from both sides
∴ m∠APD = 30°
∵ m∠DAP = m∠APD = 30°
- If two angles in a triangle are equal in measures, then the triangle
is isosceles
∴ Δ ADP is an isosceles triangle
∴ AD = DP
In Δ BPC:
∵ m∠PBC = 60° ⇒ proved
∵ m∠C = 60° ⇒ proved
- Sum of the measures of the interior angles in a Δ is 180°
∴ m∠BPC + m∠PBC + m∠C = 180°
∴ m∠BPC + 60 + 60 = 180
- Add like terms in the left hand side
∴ m∠BPC + 120 = 180
- Subtract 120 from both sides
∴ m∠BPC = 60°
∵ m∠PBC = m∠C = m∠BPC = 60°
- If the three angles of a triangle are equal in measure, then
the triangle is equilateral
∴ Δ BPC is an equilateral triangle
∴ PB = PC = BC
(iii)
∵ AD = BC ⇒ opposite sides in parallelogram
∵ AD = DP ⇒ Proved
- Equate the two right hand sides of AD
∴ BC = DP
∵ BC = PC
- Equate the right hand sides of BC
∴ DP = PC
∵ DC = DP + PC
∵ DP = AD
∴ PC = AD
- Substitute DP by AD and PC by AD in CD
∴ CD = AD + AD
∴ DC = 2 AD
Learn more:
You can learn more about parallelogram in brainly.com/question/6779145
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