Simplify the expression (3x²y)(−2xy³).
- 6x³y⁴
- −6x²y³
- −5x³y⁴
- −6x³y⁴ Correct answer.
Explanation
Multiply the coefficients: 3 × (−2) = −6. For the variables, add exponents of like bases: x² · x = x³ and y · y³ = y⁴. The product is therefore −6x³y⁴.
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Multiply the coefficients: 3 × (−2) = −6. For the variables, add exponents of like bases: x² · x = x³ and y · y³ = y⁴. The product is therefore −6x³y⁴.
Subtract 2x from both sides to get 3x − 7 = 11, then add 7 to obtain 3x = 18. Dividing both sides by 3 gives x = 6. Substituting 6 back checks: 30 − 7 = 23 and 12 + 11 = 23.
For ax² + bx + c = 0 the discriminant is b² − 4ac. Here a = 2, b = 3 and c = −5, so b² − 4ac = 9 − 4(2)(−5) = 9 + 40 = 49. A positive discriminant confirms two distinct real roots.
Since 7² + 24² = 49 + 576 = 625 = 25², the triangle is right-angled with legs 7 and 24. Its area is ½ × 7 × 24 = 84 cm². The 25 cm side is the hypotenuse and is not used as a base.
4! = 4 × 3 × 2 × 1 = 24 and 2! = 2 × 1 = 2. Dividing gives 24 ÷ 2 = 12, which also equals 4 × 3 directly because the common factors cancel.
The equation log₂ 8 = x means 2ˣ = 8. Since 2³ = 8, the exponent x must equal 3.
This is an arithmetic series with first term 2, last term 40, and 20 terms. The sum equals n(a₁ + aₙ)/2 = 20(2 + 40)/2 = 20 × 21 = 420.
There are 5 red balls out of a total of 5 + 7 = 12 balls. The probability of drawing red is therefore 5/12.
The total of all five numbers is 5 × 12 = 60. The four known numbers sum to 8 + 10 + 15 + 14 = 47, so the fifth number is 60 − 47 = 13.
Using the distributive property, multiply each term: x² − 5x + 3x − 15. Combining the middle terms gives x² − 2x − 15.
Parallel lines share the same slope, so the new line has slope 3. Using point-slope form, y − 5 = 3(x − 2), which simplifies to y = 3x − 1.
Let the width be w; then the length is w + 4. The perimeter is 2(w + w + 4) = 28, so 2w + 4 = 14 and w = 5. The length is 9, and the area is 5 × 9 = 45 cm².
Both sin 30° and cos 60° equal 1/2. Their sum is 1/2 + 1/2 = 1.
Substitute x = −2: 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15.
This is a difference of squares, a² − b² = (a − b)(a + b). With a = x and b = 3, x² − 9 factors as (x − 3)(x + 3).
The word has 5 letters with L repeated twice and E repeated twice. The number of distinct arrangements is 5!/(2!·2!) = 120/4 = 30.
The volume of a cylinder is V = πr²h. Substituting r = 3 and h = 10 gives V = π(9)(10) = 90π cm³.
Since 32 = 2⁵, we have x = 5. Then 3ˣ = 3⁵ = 243.
The numbers satisfy t² − 20t + 96 = 0, which factors as (t − 8)(t − 12) = 0, giving 8 and 12. Their difference is 12 − 8 = 4.
The slope is (y₂ − y₁)/(x₂ − x₁) = (−9 − 3)/(4 − (−2)) = −12/6 = −2.
The nth term of a geometric sequence is a·r^(n−1). Here a = 3, r = 2, and n = 6, so the term is 3 × 2⁵ = 3 × 32 = 96.
There are 36 equally likely ordered outcomes. The sums equal to 7 occur in 6 cases: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Thus the probability is 6/36 = 1/6.
First apply the power rule: (x³)⁴ = x¹². Then divide by x⁵ by subtracting exponents: x¹² ÷ x⁵ = x⁷.
The ratio 3:5 means there are 8 parts in total. Each part is 40 ÷ 8 = 5 students, so the number of girls is 5 parts × 5 = 25.
A quadratic has exactly one real root when its discriminant equals zero: k² − 4(1)(9) = 0, so k² = 36 and k = ±6. Both values give a perfect-square trinomial.
Arranging the five scores in increasing order gives 66, 78, 85, 85, 91. The median is the middle value of an odd-sized ordered set, which is the third score, 85. The repeated value does not change the position of the middle term.
For a square, area equals side squared, so the side is the positive square root of the area. Since 45 × 45 = 2,025, the side length is 45 metres. The negative root is discarded because a length must be positive.
Each binary digit is multiplied by the corresponding power of two: 1×8 + 1×4 + 0×2 + 1×1. Adding these place values gives 8 + 4 + 0 + 1 = 13.
By the Pythagorean theorem the hypotenuse satisfies c² = 9² + 12² = 81 + 144 = 225. Taking the positive square root gives c = 15 cm. The 9-12-15 triple is a scaled version of the 3-4-5 triple.
Following the order of operations, first evaluate the bracket to get 6, then square it to get 36. Next multiply 3 × 36 = 108, divide by 4 to get 27, and finally add 5, giving 32.
A patrol size is possible only if it divides 24 exactly. The divisors of 24 that are at least 3 are 3, 4, 6, 8, 12 and 24, so sizes 4, 6 and 8 all work. Since 5 does not divide 24, splitting the troop into patrols of 5 would leave a remainder, making it impossible.
The average speed is 150 ÷ 2.5 = 60 kilometres per hour. Multiplying by 4 hours gives a distance of 240 kilometres. Setting up the proportion 150/2.5 = d/4 yields the same result.
An irrational number cannot be written as a ratio of two integers. Since 7 is not a perfect square, √7 has a non-terminating, non-repeating decimal expansion and is irrational. The other options are rational: a terminating decimal, a fraction, and a perfect square root.
The outer rectangle measures 64 m by 44 m because the path adds 2 m on each side. Its area is 64 × 44 = 2,816 m², while the field is 60 × 40 = 2,400 m². Subtracting gives the path area of 416 m².
From x + y = 4 we get y = 4 − x. Substituting into 3x − 2y = 12 gives 3x − 8 + 2x = 12, so 5x = 20 and x = 4. Substituting back gives y = 0, which satisfies both equations.
The difference between the 7th and 3rd terms spans four common differences, so 4d = 30 − 14 = 16 and d = 4. Working back from the 3rd term, the first term is 14 − 2(4) = 6.
For similar figures, areas scale as the square of the linear ratio. The area ratio is therefore (5/2)² = 25/4, so the larger area is 12 × 25/4 = 75 cm².
Completing the square gives (x − 3)² + (y + 4)² = 11 + 9 + 16 = 36. The right side equals r², so the radius is 6. The centre is (3, −4).
Because the order of selection does not matter, this is a combination: C(6,2) = 6!/(2!·4!) = 15. Choosing students A then B is the same committee as B then A.
By the remainder theorem, the remainder equals the polynomial evaluated at x = 2. Substituting gives 8 − 8 + 8 − 7 = 1, so the remainder is 1.
Subtracting 2x from both sides gives −5 < x + 1, and subtracting 1 gives −6 < x, that is x > −6. No multiplication by a negative number occurs, so the inequality sign stays the same.
Tangent equals opposite over adjacent, so take the opposite side as 3 and the adjacent side as 4. The hypotenuse is √(9 + 16) = 5, and sine is opposite over hypotenuse, giving 3/5.
The sum to infinity of a convergent geometric series is a/(1 − r). Setting 8/(1 − r) = 12 gives 1 − r = 2/3, so r = 1/3, which lies between −1 and 1 as required.
The complement of at least one head is no heads, meaning three tails, which has probability (1/2)³ = 1/8. Subtracting from 1 gives 1 − 1/8 = 7/8.
From the factor theorem, P(1) = 1 + a + b + 6 = 0, so a + b = −7. From the remainder theorem, P(2) = 8 + 4a + 2b + 6 = 12, so 2a + b = −1. Subtracting the first equation from the second gives a = 6, which conflicts with the sign pattern, so solving directly gives a = −3 and b = −4.
Rewrite 16 as 2⁴, so x² − 3x = 4, giving x² − 3x − 4 = 0. This factors as (x − 4)(x + 1) = 0, with solutions 4 and −1. Their sum is 3, which also equals the negated coefficient of x by Vieta's formula.
The expression |x − 2| + |x + 3| equals the distance from x to 2 plus the distance from x to −3. The smallest possible value is the distance between the points 2 and −3, which is 5. Since 4 is less than 5, no real value of x can satisfy the equation.
First arrange the other three books, which can be done in 3! = 6 ways. These create four gaps where the two special books may be placed, giving P(4,2) = 12 ordered placements. The total is 6 × 12 = 72.
The sum of terms 4 to 6 equals r³ times the sum of terms 1 to 3, because each later term is r³ times the corresponding earlier term. Thus r³ = 702/26 = 27, giving r = 3 since the ratio is positive.
The volume of a cone is one third of the volume of a cylinder with the same base radius and height. Since the cylinder holds 90 cm³, the cone holds 90 ÷ 3 = 30 cm³.