Simplify the expression 3(2x − 4) + 5x.
- 6x − 7
- 11x − 12 Correct answer.
- 6x − 12
- 11x − 4
Explanation
Distribute the 3 to get 6x − 12, then add 5x to combine like terms, giving 11x − 12. The constant term stays −12 because no other constant is added.
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Distribute the 3 to get 6x − 12, then add 5x to combine like terms, giving 11x − 12. The constant term stays −12 because no other constant is added.
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The sum is 3 + 7 + 7 + 9 + 14 = 40, and there are 5 values, so the mean is 40 ÷ 5 = 8. The mean need not equal any individual value or the median.
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Substitute x = −2: 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15. Squaring the negative gives a positive 4, and the −3 times −2 becomes +6.
We need two numbers that multiply to +6 and add to −5, namely −2 and −3. Thus x² − 5x + 6 = (x − 2)(x − 3), which expands back correctly.
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log₂ 32 asks for the exponent to which 2 must be raised to obtain 32. Since 2⁵ = 32, the value is 5.
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A quadratic has exactly one real solution when its discriminant b² − 4ac equals zero. Here k² − 36 = 0, so k = 6 or k = −6.
The sector area is (θ/360)·πr² = (72/360)·25π = (1/5)·25π = 5π. The sector is one-fifth of the full circle.
On one roll, the numbers greater than 4 are 5 and 6, so the probability is 2/6 = 1/3. Independent rolls multiply: (1/3)·(1/3) = 1/9.
The nth term of a geometric sequence is a·r^(n−1). Here the 6th term is 3·2⁵ = 3·32 = 96.
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The first term is 8 and the common ratio is 1/2, so the sum is a/(1 − r) = 8/(1 − 1/2) = 8/(1/2) = 16. The series converges because |r| < 1.
The semiperimeter is s = (7 + 8 + 13)/2 = 14. By Heron's formula, area = √(14·7·6·1) = √588 = 14√3? That is not among the answers, so recheck: 14·7·6·1 = 588, and √588 ≈ 24.25, matching 24 only approximately.
Dividing P(x) by (x − 2) gives x² − 2x − 3, which factors as (x − 3)(x + 1). Thus the roots are x = 2, x = 3, and x = −1.
Powers of i repeat with period 4, so i⁴ = 1 and the exponent reduces modulo 4. Since 47 = 4·11 + 3, i⁴⁷ = i³ = i²·i = −i.
In f(g(x)) the output of g is substituted into f, so f(2x) = 2x + 3. The inner function is applied first and then the outer function adds three to that result.
The transpose Mᵀ has three rows and two columns. A 2×3 matrix multiplied by a 3×2 matrix gives a 2×2 matrix, since the inner dimensions match and the outer dimensions give the result size.
By the inclusion–exclusion principle the number studying at least one subject is 18 + 22 − 8 = 32. Subtracting from the total gives 40 − 32 = 8 students studying neither.
The binomial coefficient is 6!/(2!·4!) = (6·5)/(2·1) = 15. Order does not matter when choosing a subset, so the permutation count 30 is halved.
The magnitude is the square root of the sum of squared components: √(3² + 4²) = √25 = 5. This is the Pythagorean length of the vector.
The power rule states that the derivative of xⁿ is n·xⁿ⁻¹. Applying it to 4x³ gives 4·3·x² = 12x².
The probability of a red first is 5/8, and after removing one red the probability of a second red is 4/7. Multiplying gives (5/8)·(4/7) = 20/56 = 5/14.
The common difference is 4, so the nth term is aₙ = 7 + 4(n − 1). For n = 30 this gives 7 + 4·29 = 7 + 116 = 123.
By the remainder theorem, the remainder equals P(2). Computing 8 + 8 − 10 + 1 gives 7.
The standard form is (x − h)² + (y − k)² = r², where (h, k) is the center. Substituting h = −2, k = 3, and r² = 16 gives (x + 2)² + (y − 3)² = 16.
Using the Pythagorean identity sin²θ + cos²θ = 1, we get cos²θ = 1 − 9/25 = 16/25. Since θ is acute, cos θ is positive, so cos θ = 4/5.
Solving y = 3x − 6 for x gives x = (y + 6)/3. Swapping the variables produces f⁻¹(x) = (x + 6)/3.
By the binomial theorem the term containing x³ corresponds to choosing x three times and 2 twice. The coefficient is C(5,2)·2² = 10·4 = 40.
An antiderivative of 2x is x². Evaluating from 1 to 3 gives 3² − 1² = 9 − 1 = 8, which is the area under the line between those limits.
Perpendicular lines have slopes whose product is −1. The given slope is −2, so the required slope is the negative reciprocal, 1/2.
For |r| < 1 the infinite sum is a/(1 − r). Substituting a = 12 and r = 1/3 gives 12/(2/3) = 18.
The expression under a square root must be non-negative, so x − 4 ≥ 0, giving x ≥ 4. The value x = 4 is included because the square root of zero is defined.
For a matrix with entries a, b in the first row and c, d in the second, the determinant is ad − bc. Here 4·3 − 1·2 = 12 − 2 = 10.
Substituting u = x² gives u² − 5u + 4 = 0, which factors as (u − 1)(u − 4) = 0. Thus x² = 1 or x² = 4, giving roots ±1 and ±2, so the remaining roots are 2 and −2.
There are 2⁴ = 16 equally likely sequences, and C(4,2) = 6 of them have exactly two heads. The probability is 6/16 = 3/8.
The derivative 3x² − 12x + 9 factors as 3(x − 1)(x − 3), giving critical points at x = 1 and x = 3, with f(1) = 4 and f(3) = 0. Checking endpoints gives f(0) = 0 and f(4) = 4, so the maximum value is 4.
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Using the shoelace formula with the vertices, the area is ½|x_A(y_B − y_C) + x_B(y_C − y_A) + x_C(y_A − y_B)|. This gives ½|0 + 6·4 + 2·0| = ½·24 = 12.