idn34 ef djewfe ferkflew?

idn34 ef djewfe ferkflew?
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1 Answer

Answer :

fiwjnefwkfew ;(
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Related questions

There is an anime, "Ef: A Fairy Tale of the Two", which puts in evidence a think that I would not expect. Am I right on this one?

Description : There is an anime, "Ef: A Fairy Tale of the Two", which puts in evidence a think that I would not expect. Am I right on this one?

Last Answer : How are either of them to blame for her losing her eye in an accident when neither of them were present?

Does anyone have any experience with EF Tours?

Description : Does anyone have any experience with EF Tours?

Last Answer : answer:@AstroChuck Long time no see! My daughter went to Costa Rica a couple years ago on a class trip, through EF Tours. For the most part we were pretty satisfied with the service and what she got ... card that we could follow and control through a US bank. (Actually, I think it was from USBank).

I have a very generic carry-on bag from EF Tours for my trip, ideas on personalizing it?

Description : I have a very generic carry-on bag from EF Tours for my trip, ideas on personalizing it?

Last Answer : I would try fabric paint.

What's the difference between EF and EF-S designation for Canon lenses?

Description : What's the difference between EF and EF-S designation for Canon lenses?

Last Answer : EF means that it is an electronic focus lens, basically every lens made since the late 80s. Also called auto focus. EF-S are crop sensor camera lenses. They will only fit on APS-C cameras, which ... L lens you know you are getting the best optical quality available. You will also pay dearly for it.

I'm looking for a refurbished Canon EF 50mm f/1.4 USM Lens in Canada. Where can I get one?

Description : I'm looking for a refurbished Canon EF 50mm f/1.4 USM Lens in Canada. Where can I get one?

Last Answer : Ebay or BestBuy.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th

Description : In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th

Last Answer : Solution :-

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Last Answer : Solution :-

A square current carrying loop abcd is palced near an infinitely long another current carrying wire ef. Now, match the following two columns.

Description : A square current carrying loop abcd is palced near an infinitely long another current carrying wire ef. Now, match the following two columns.

Last Answer : A square current carrying loop abcd is palced near an infinitely long another current carrying wire ef. Now, match the following two columns.

How can the IMA of a first- class lever be increased Decrease the length between the applied effort and the pivot. Increase the length between the applied effort and the pivot. Decrease the applied ef?

Description : How can the IMA of a first- class lever be increased Decrease the length between the applied effort and the pivot. Increase the length between the applied effort and the pivot. Decrease the applied ef?

Last Answer : What is the answer ?

Diphtheria toxin inhibits (A) Prokaryotic EF-1 (B) Prokaryotic EF-2 (C) Eukaryotic EF-1 (D) Eukaryotic EF-2

Description : Diphtheria toxin inhibits (A) Prokaryotic EF-1 (B) Prokaryotic EF-2 (C) Eukaryotic EF-1 (D) Eukaryotic EF-2

Last Answer : Answer : D

Translocation of the newly formed peptidyl tRNA at the A site into the empty P site involves (A) EF-II, GTP (B) EF-I, GTP (C) EF-I, GDP (D) Peptidyl transferase, GTP

Description : Translocation of the newly formed peptidyl tRNA at the A site into the empty P site involves (A) EF-II, GTP (B) EF-I, GTP (C) EF-I, GDP (D) Peptidyl transferase, GTP

Last Answer : Answer : A

The newly entering amino acyl tRNA into A site requires (A) EF-II (B) Ribosomal RNA (C) mRNA (D) EF-I

Description : The newly entering amino acyl tRNA into A site requires (A) EF-II (B) Ribosomal RNA (C) mRNA (D) EF-I

Last Answer : Answer : D

In the process of elongation of chain binding of amino acyl tRNA to the A site requires (A) A proper codon recognition (B) GTP (C) EF-II (D) GDP

Description : In the process of elongation of chain binding of amino acyl tRNA to the A site requires (A) A proper codon recognition (B) GTP (C) EF-II (D) GDP

Last Answer : Answer : A

The formation of initiation complex during protein synthesis requires a factor: (A) IF-III (B) EF-I (C) EF-II (D) IF-I

Description : The formation of initiation complex during protein synthesis requires a factor: (A) IF-III (B) EF-I (C) EF-II (D) IF-I

Last Answer : Answer : A

To calculate early start (ES) and early finish (EF) perform a: A. Backward pass B. Forward pass C. Lateral pass D. Critical pass

Description : To calculate early start (ES) and early finish (EF) perform a:  A. Backward pass B. Forward pass  C. Lateral pass D. Critical pass

Last Answer : B. Forward pass

In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then (a) DE || BC (b) DF || AC (c) EF || AB (d) none of

Description : In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then (a) DE || BC (b) DF || AC (c) EF || AB (d) none of

Last Answer : (c) EF || AB

If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF then which of the following is not true? (a) BC.EF = AC.FD (b) AB.EF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD

Description : If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF then which of the following is not true? (a) BC.EF = AC.FD (b) AB.EF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD

Last Answer : (c) BC.DE = AB.EF

If D is the duration, ES and EF are the earliest start and finish, LS and LF are latest start and latest finish time, then the following relation holds good (A) EF = ES + D (B) LS = LF - D (C) LF = LS + D (D) All the above

Description : If D is the duration, ES and EF are the earliest start and finish, LS and LF are latest start and latest finish time, then the following relation holds good (A) EF = ES + D (B) LS = LF - D (C) LF = LS + D (D) All the above

Last Answer : (D) All the above

The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Description : The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Last Answer : 111 psi

The result of adding hexadecimal number A6 to 3A is (A) DD. (B) E0. (C) F0. (D) EF.

Description : The result of adding hexadecimal number A6 to 3A is (A) DD. (B) E0. (C) F0. (D) EF.

Last Answer : (B) E0.

How many two-input AND and OR gates are required to realize Y=CD+EF+G (A) 2,2. (B) 2,3. (C) 3,3. (D) none of these.

Description : How many two-input AND and OR gates are required to realize Y=CD+EF+G (A) 2,2. (B) 2,3. (C) 3,3. (D) none of these.

Last Answer : Ans: A Y=CD+EF+G Number of two input AND gates=2 Number of two input OR gates = 2 One OR gate to OR CD and EF and next to OR of G & output of first OR gate.

How many AND gates are required to realize Y = CD+EF+G (A) 4 (B) 5 (C) 3 (D) 2

Description : How many AND gates are required to realize Y = CD+EF+G (A) 4 (B) 5 (C) 3 (D) 2

Last Answer : 2

Botanical Words Alphabetical List - EF

Description : Botanical Words Alphabetical List - EF

Last Answer : EFFLORESCENCE:The time or state of flowering; anthesis. EFFLORESCENT: Blooming or flowering.

Find the back bearing of the following lines having fore bearing as given below: (i) PQ = N 55° 0' E (ii) EF == 280° 0' (iii) CD = S 58° 30' W (iv) OM= 180° 0'

Description : Find the back bearing of the following lines having fore bearing as given below: (i) PQ = N 55° 0' E (ii) EF == 280° 0' (iii) CD = S 58° 30' W (iv) OM= 180° 0'

Last Answer : i) F.B OF PQ = N 55°0 ’ E  B.B Of PQ = S 55°0 ’W  ii) F.B. Of EF = 280°0 ’   B.B OF EF = F.B -180°= 280°0 ’ -180° = 100°0 ’  iii) F.B OF CD = S 58°30’W  B.B OF CD = N 58°30’ E  iv) F.B OF GH = 180°  B.B OF GH = F.B -180°=180°-180°= 0°

The reverse polish notation equivalent to the infix expression ((A + B) * C + D)/(E + F + G) (A) A B + C * D + EF + G + / (B) A B + C D * + E F + G + / (C) A B + C * D + E F G + +/ (D) A B + C * D + E + F G + /

Description : The reverse polish notation equivalent to the infix expression ((A + B) * C + D)/(E + F + G) (A) A B + C * D + EF + G + / (B) A B + C D * + E F + G + / (C) A B + C * D + E F G + +/ (D) A B + C * D + E + F G + /

Last Answer : (A) A B + C * D + EF + G + /

Convert the following infix expression into its equivalent post fix expression (A + B^ D) / (E – F) + G (A) ABD^ + EF – / G+ (B) ABD + ^EF – / G+ (C) ABD + ^EF / – G+ (D) ABD^ + EF / – G+

Description : Convert the following infix expression into its equivalent post fix expression (A + B^ D) / (E – F) + G (A) ABD^ + EF – / G+ (B) ABD + ^EF – / G+ (C) ABD + ^EF / – G+ (D) ABD^ + EF / – G+

Last Answer : (A + B^ D) / (E – F) + G

Let R be the rectangular window against which the lines are to be clipped using 2D Sutherland-Cohen line clipping algorithm. The rectangular window has lower left-hand corner at (-5,1) and upper righthand corner at (3,7). Consider the following three lines for clipping with the given end point co-ordinates: Line AB: A(-6,2) and B(-1,8) Line CD: C(-1,5) and D(4,8) Line EF: E(-2,3) and F(1,2) Which of the following line(s) is/are candidate for clipping? (A) AB (B) CD (C) EF (D) AB and CD

Description : Let R be the rectangular window against which the lines are to be clipped using 2D Sutherland-Cohen line clipping algorithm. The rectangular window has lower left-hand corner at (-5,1) and upper righthand corner at (3,7). ... s) is/are candidate for clipping? (A) AB (B) CD (C) EF (D) AB and CD

Last Answer : (D) AB and CD

In an n-type semiconductor, as the donor concentration ND increases, the Fermi level EF:

Description : In an n-type semiconductor, as the donor concentration ND increases, the Fermi level EF: (1) Remains unaltered (2) Moves towards the conduction band (3) Move towards the center of forbidden energy gap (4) May or may not move depending on temperature 

Last Answer : In an n-type semiconductor, as the donor concentration ND increases, the Fermi level EF: Moves towards the conduction band