What is the value of the discriminant of the quadratic equation 2x² − 5x + 3 = 0?
- 25
- 1 Correct answer.
- 49
- −1
Explanation
For ax² + bx + c = 0 the discriminant is b² − 4ac. Here a = 2, b = −5, c = 3, so (−5)² − 4·2·3 = 25 − 24 = 1.
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For ax² + bx + c = 0 the discriminant is b² − 4ac. Here a = 2, b = −5, c = 3, so (−5)² − 4·2·3 = 25 − 24 = 1.
When multiplying powers with the same base, add the exponents: x³·x² = x⁵ and y²·y⁵ = y⁷. The product is therefore x⁵y⁷.
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The interior angles of a convex n-gon sum to (n − 2)·180°. For n = 8 this gives 6·180° = 1080°.
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log₂ 32 asks for the exponent e with 2ᵉ = 32. Since 2⁵ = 32, the value is 5.
The nth term of an arithmetic sequence is aₙ = a₁ + (n − 1)d. Thus a₁₀ = 7 + 9·4 = 7 + 36 = 43.
The Pythagorean identity states that sin²θ + cos²θ = 1 for all θ. It follows from the coordinates of a point on the unit circle.
Subtract 3 from both sides to get −2x < 4. Dividing by −2 reverses the inequality, giving x > −2.
Equal roots require a zero discriminant: b² − 4ac = 36 − 4k = 0. Solving gives k = 9, and indeed x² − 6x + 9 = (x − 3)².
In a geometric sequence aₙ = a₁rⁿ⁻¹, so 24 = 3r³, giving r³ = 8 and r = 2.
Differentiate term by term using the power rule: the derivative of x³ is 3x² and the derivative of −4x is −4. The result is 3x² − 4.
Perpendicular vectors have zero dot product: 3k + (−1)(2) = 0, so 3k = 2 and k = 2/3.
Order does not matter, so this is a combination: C(8,3) = 8!/(3!·5!) = (8·7·6)/(3·2·1) = 56.
For |r| < 1 the sum is a/(1 − r). Here a = 1 and r = 1/2, so the sum is 1/(1 − 1/2) = 2.
The slope is (11 − 2)/(4 − 1) = 9/3 = 3. Using point (1, 2): y − 2 = 3(x − 1), so y = 3x − 1.
Use sin²θ = 1 − cos²θ = 1 − 9/25 = 16/25. In the first quadrant sin θ is positive, so sin θ = 4/5.
An antiderivative is x³ + x. Evaluating from 0 to 2 gives (8 + 2) − 0 = 10.
Complete the square: x² − 4x + 7 = (x − 2)² + 3. The square is never negative, so the minimum is 3, attained at x = 2.
Write 8 as 2³, so 8^(x−1) = 2^(3x−3). Equating exponents: x + 1 = 3x − 3, giving 4 = 2x and x = 2.
Dividing by (x − 1) gives x² − x − 6, which factors as (x − 3)(x + 2). So the roots are 1, 3 and −2.
Complete the square: (x − 3)² + (y + 4)² = 11 + 9 + 16 = 36. The radius is √36 = 6.
By the cosine rule, BC² = 7² + 9² − 2·7·9·cos 60° = 49 + 81 − 63 = 67, so BC = √67.
The probability the first is red is 3/5; given that, the second is red with probability 2/4. Multiplying gives (3/5)(2/4) = 6/20 = 3/10.
This is geometric with ratio r = 2x. A geometric series converges exactly when |r| < 1, so |2x| < 1 and hence |x| < 1/2.
Since 6² + 8² = 36 + 64 = 100 = 10², the triangle is right-angled with legs 6 and 8. Its area is half the product of the legs, (6 × 8)/2 = 24.
The sum of the first n odd integers equals n², since the sequence is arithmetic with a₁ = 1 and d = 2. Here n = 20, so the sum is 20² = 400.
Each entry of Ax is the dot product of a row of A with x. First row: 2·1 + 1·2 = 4; second row: 1·1 + 3·2 = 7, giving [[4], [7]].
By the remainder theorem, the remainder equals the value of the polynomial at x = 2. Substituting gives 8 + 8 − 10 + 1 = 7.
The word has 5 letters with L repeated twice and E repeated twice, so the number of distinct arrangements is 5!/(2!·2!) = 120/4 = 30.
The double-angle identity states that 2 sin θ cos θ = sin 2θ. Taking θ = 15° gives sin 30° = 1/2.
Setting f′(x) = 3x² − 6x = 3x(x − 2) to zero gives critical points x = 0 and x = 2. The second derivative f″(x) = 6x − 6 is negative at x = 0, so f has a local maximum there.
Mongolian vowels are divided into back vowels (а, о, у) and front vowels (э, ө, ү). The vowel 'у' belongs to the back vowel set, while 'ү', 'ө', and 'э' are front vowels.
The plural suffix -ууд is attached to back-vowel words ending in consonants, so 'ном' becomes 'номууд'. The form 'номүүд' would violate vowel harmony, and 'номнар' and 'номуд' are not standard plural forms.
The accusative case marks the direct object of a transitive verb and is formed with the suffix -ыг/-ийг. In 'Би номыг уншив' (I read the book), 'номыг' correctly marks the object. The other suffixes indicate location, ablative, or instrumental meanings, which do not fit the verb 'унших' (to read) with a direct object.
A compound word is formed by combining two or more independent words into a single lexical unit. 'Гэр бүл' (family) is composed of 'гэр' (home) and 'бүл' (group), making it a compound. The other options are simple, underived words.
Synonyms are words with similar meanings. 'Үзэсгэлэнтэй' means 'beautiful' or 'gorgeous', closely matching 'сайхан'. The other options mean 'ugly', 'small', and 'big', which are not synonyms.
Antonyms are words with opposite meanings. The opposite of 'өндөр' (high/tall) is 'нам' (low). 'Өргөн' means 'wide', 'хол' means 'far', and 'гүн' means 'deep', none of which are direct opposites of height.
Mongolian basic word order is subject–object–verb (SOV), with time expressions typically placed before the object or at the beginning. 'Би өнөөдөр сургуульд явна' (I today to school go) follows the natural SOV order with the time adverb before the object. The other options place the verb or elements in unnatural positions.
The simple past tense in Mongolian is formed with the suffix -лаа/-лээ/-лоо/-лөө, depending on vowel harmony. For 'явах', the stem 'яв-' takes '-лаа' to make 'явлаа' (went). The suffix '-на' marks present-future, '-ж байна' marks progressive, and '-х' marks future or verbal noun.
'Маш' means 'very' and modifies the adjective 'өндөр' (tall), specifying degree. Words that modify adjectives or verbs are adverbs. It is not an adjective, noun, or verb.
A declarative sentence makes a statement and typically ends with a period. 'Би маргааш ирнэ.' (I will come tomorrow.) is a statement. The others are interrogative (questions) or imperative (command).
Interrogative sentences in Mongolian, as in many languages, end with a question mark (?). A period ends declarative sentences, an exclamation mark ends exclamatory or imperative sentences, and a comma separates elements within a sentence.
'Жангар' is a traditional Mongolian epic that celebrates the heroic deeds of the warrior Jangar and his companions as they defend their land. It is not primarily about a monk, a romance, or daily nomadic life.
'Газар шороо' is a novel by the prominent Mongolian writer Бямбын Ринчен. It is a classic work of Mongolian literature. The other authors are also well-known but did not write this novel.
The present progressive in Mongolian is formed with the converb -ж plus the auxiliary 'байна'. Thus 'уншиж байна' means 'am reading'. The other forms represent present-future, past, and future/verbal noun respectively.
A simile compares two things using words like 'шиг' (like/as). 'Түүний нүд од шиг гялалзана' (Her eyes shine like stars) uses 'шиг' to compare eyes to stars. The other sentences are literal statements without comparison.
The traditional Mongolian four-line stanza 'дөрвөн эгнээний шүлэг' typically uses alliteration and end-rhyme with an AAAA scheme, where all four lines rhyme. Other schemes like AABB or ABAB are less common in this specific form.
'Миний нутаг' is a well-known poem by Дашдоржийн Нацагдорж, a foundational figure in modern Mongolian literature. The other poets are also significant but did not write this specific poem.
The suffix '-ын/-ийн' marks the genitive case, indicating possession or relation, as in 'багшийн' (of the teacher). It is not a plural, accusative, or tense marker.