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Qormaata galuumsa kutaa 12ffaa Herregaa, bara 2018 A.L.I (2026), maxxansa 1

Gaaffilee dhugaa qormaata galuumsa yunivarsiitii kutaa 12 kan Herregaa, akkuma barattoonni bara 2018 A.L.I itti fudhatanitti. Tokkoon tokkoon gaaffii deebii sirrii fi ibsa qaba. Qormaanni bara 2018 A.L.I maxxansa adda addaatiin bahe. Fuulli kun maxxansa 1 qaba; kaan gadi aanee tarreeffamaniiru.

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  • 2018 A.L.I (2026), maxxansa 1

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  • Gaaffilee 29 deebii waliin

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Gaaffilee qormaata kanaa yaalaa

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Gaaffii 3

After rationalizing the expression 3/(3√3 - 2√2), the denominator is equal to

  1. A.19
  2. B.57
  3. C.3(3√3 - 2√2)/19
  4. D.3(3√3 + 2√2)/19
Deebii fi ibsa ilaalaa

Deebii: A 19

Maaliif

To rationalize, multiply the top and the bottom by the conjugate of the denominator, 33+223\sqrt{3}+2\sqrt{2}. The bottom then becomes a difference of two squares: (33)2−(22)2=9×3−4×2=27−8=19(3\sqrt{3})^2-(2\sqrt{2})^2 = 9\times 3 - 4\times 2 = 27-8 = 19. So the denominator the question asks for is 1919. Option C is tempting because it is the value of the whole fraction after rationalizing, but the question asks only for what sits underneath.

Gaaffii 5

The solution set of the system {3x - y = 5, 3y + 15 = 9x} is

  1. A.{(0,-5),(1,-2),(2,-1),...}
  2. B.R
  3. C.∅
  4. D.{(t,3t-5): t∈R}
Deebii fi ibsa ilaalaa

Deebii: D {(t,3t-5): t∈R}

Maaliif

The first equation gives y=3x−5y = 3x-5. Rearrange the second: 3y=9x−153y = 9x-15, so y=3x−5y = 3x-5 as well. The two equations describe the same line, so every point on that line solves the system. Writing xx as any real number tt, the solutions are the pairs (t, 3t−5)(t,\,3t-5). Option B is tempting because there really are infinitely many solutions, but those solutions are points on one line, not every point of the plane.

Gaaffii 6

The solution set of |2x+1| = x-7 is

  1. A.{-8,-2}
  2. B.{-8,2}
  3. C.{8,2}
  4. D.{}
Deebii fi ibsa ilaalaa

Deebii: D {}

Maaliif

An absolute value is never negative, so ∣2x+1∣=x−7|2x+1| = x-7 can only work when x−7≥0x-7 \ge 0, that is when x≥7x \ge 7. Take the two cases anyway: 2x+1=x−72x+1 = x-7 gives x=−8x = -8, and 2x+1=−(x−7)2x+1 = -(x-7) gives 3x=63x = 6, so x=2x = 2. Neither −8-8 nor 22 reaches 77, so both fail the check and the solution set is empty. Option B collects those two candidates, but a candidate only becomes a solution after it survives substitution.

Gaaffii 8

Suppose N is the set of natural numbers and A = {1,2,{3}}, then which of the following is true?

  1. A.{3} ⊆ A
  2. B.N ∪ A = N
  3. C.A\{3} = A
  4. D.A ⊆ N
Deebii fi ibsa ilaalaa

Deebii: C A\{3} = A

Maaliif

In A={1, 2, {3}}A = \{1,\,2,\,\{3\}\} the third element is the set {3}\{3\}, not the number 33. Since 33 itself is not an element of AA, taking it away removes nothing, so A∖{3}=AA \setminus \{3\} = A. Option A says {3}⊆A\{3\} \subseteq A, which would need the number 33 to sit inside AA. Here {3}\{3\} is a member of AA, and being a member is a different thing from being a subset.

Gaaffii 10

Let A={0,1,2,3,4} and R be a relation from A to A given by R={(x,y): y=x+2, for x,y∈A}. Which one is true about R?

  1. A.(2,3) is an element of R
  2. B.(3,5) is an element of R
  3. C.The domain of R is {1,2,3,4}
  4. D.The range of R is {2,3,4}
Deebii fi ibsa ilaalaa

Deebii: D The range of R is {2,3,4}

Maaliif

The rule is y=x+2y = x+2, with both xx and yy taken from A={0,1,2,3,4}A=\{0,1,2,3,4\}. Only x=0,1,2x=0,1,2 keep yy inside AA, so R={(0,2),(1,3),(2,4)}R = \{(0,2),(1,3),(2,4)\}. The second coordinates are 2,3,42, 3, 4, which makes the range {2,3,4}\{2,3,4\}. Option B looks right if you forget that yy must also come from AA: 55 is not in AA, so (3,5)(3,5) never enters RR.

Gaaffii 12

The graph of f(x)=x²+6x+7 is obtained from the graph of g(x)=x² by shifting

  1. A.6 units to the right and 9 units upward
  2. B.3 units to the right and 2 units downward
  3. C.3 units to the left and 2 units downward
  4. D.3 units to the left and 2 units upward
Deebii fi ibsa ilaalaa

Deebii: C 3 units to the left and 2 units downward

Maaliif

Complete the square: x2+6x+7=(x2+6x+9)−9+7=(x+3)2−2x^2+6x+7 = (x^2+6x+9)-9+7 = (x+3)^2-2. When x2x^2 becomes (x+3)2(x+3)^2, the graph has moved 3 units to the left, because the +3+3 sits inside the bracket with xx. The −2-2 outside the bracket then pulls the whole curve 2 units down. Option B has the right sizes but the wrong horizontal direction: (x+3)2(x+3)^2 shifts left, while (x−3)2(x-3)^2 would shift right.

Qormaata guutuu PDF dhaan buufadhaa

Gaaffilee 29 kuusaa kana keessa jiran hunda akka waraqaa qormaataa maxxanfameetti, furtuu deebii ibsa 29 waliin immoo akka faayilii lammaffaatti. Maxxansaa yookiin daataa malee dubbisaa.

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Qormaatilee Herregaa biroo bara 2018 A.L.I

2018 A.L.I (2026)2018 A.L.I (2026), damee hawaasaa2018 A.L.I (2026), maxxansa 22018 A.L.I (2026), maxxansa 32018 A.L.I (2026), maxxansa 42018 A.L.I (2026), maxxansa 52018 A.L.I (2026), maxxansa 6

Qormaatilee Herregaa kan waggoota biroo

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Qormaatilee kana fakkaatan dabalataa

Qormaatilee Herregaa kutaa 12ffaa hundaQormaatilee bara 2018 A.L.I hundaGosa barnootaa fi bara hunda

Qormaata kana taa'uu dura Herregaa qayyabadhu

Yaadannoon gabaabaan Herregaa kutaa 12 kaardii 32 qaba. Tokkoon tokkoon boqonnaa boqonnaadhaan kitaaba biyyaalessaa irraa qophaa'ee seericha, mallattoowwan isaa maal bakka akka bu'an fi dogoggora qabxii nama dhabsiisu ni ibsa.

Yaadannoo Herregaa bani
Kunneen gaaffilee dhugaa qormaata galuumsa Herregaa bara 2018 A.L.I ti?

Eeyyee. Qormaata galuumsa yunivarsiitii biyyaalessaa kan Herregaa barattoonni bara 2018 A.L.I (2026) fudhatan irraa fudhataman. Kuusaan keenya gaaffilee isaa 29 qaba; tokkoon tokkoon isaanii deebii fi ibsa gara kitaaba barnootaatti qajeelchu qabu.

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