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Subtracting 2x from both sides leaves x − 7 = 5, and adding 7 to both sides gives x = 12. Checking in the original equation: 3(12) − 7 = 29 and 2(12) + 5 = 29, so the solution is consistent.
In a 30°–60°–90° triangle the sides are in the ratio 1 : √3 : 2, so the side opposite the 30° angle divided by the hypotenuse is 1/2. Therefore sin 30° equals 0.5 exactly.
The area of a circle is A = πr². Substituting r = 5 gives A = π(5)² = 25π cm².
By definition i² = −1, so i⁴ = (i²)² = (−1)² = 1. The integer powers of i therefore repeat in the cycle i, −1, −i, 1.
The sum of the first n positive integers is n(n + 1)/2. With n = 10 this gives 10 × 11 / 2 = 55.
The power rule states that d/dx (xⁿ) = n·xⁿ⁻¹. Applying it with n = 3 gives 3x².
Slope is the change in y divided by the change in x. Here (6 − 2)/(3 − 1) = 4/2 = 2.
For a 2×2 matrix with entries a, b in the first row and c, d in the second, the determinant is ad − bc. Here it is (2)(4) − (3)(1) = 8 − 3 = 5.
Factoring the quadratic gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. These roots sum to 5 and multiply to 6, matching the coefficients of the equation.
The Pythagorean identity gives cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5. In the first quadrant the cosine is positive, hence cos θ = 4/5.
The power rule fails for the exponent −1, and the antiderivative of x⁻¹ is the natural logarithm. Hence ∫ (1/x) dx = ln|x| + C, with the absolute value keeping the result valid for negative x.
The dot product is the sum of the products of corresponding components: (1)(4) + (2)(−5) + (3)(6) = 4 − 10 + 18 = 12.
Separating the variables gives dy/y = dx, and integrating yields ln y = x + C, so y = A·eˣ. The condition y(0) = 1 forces A = 1, giving y = eˣ.
The first term is a = 4 and the common ratio is r = 1/2. Since |r| < 1 the series converges to a/(1 − r) = 4/(1 − 1/2) = 8.
Because the order of selection is irrelevant, the count is the combination C(8,3) = 8!/(3!·5!). This equals (8 × 7 × 6)/(3 × 2 × 1) = 56.
For z = a + bi the modulus is √(a² + b²). Here |z| = √(3² + 4²) = √(9 + 16) = √25 = 5.
The arithmetic mean is the sum of the observations divided by the number of observations. The sum is 2 + 4 + 4 + 6 + 9 = 25 and there are 5 values, so the mean is 25/5 = 5.
By the Pythagorean theorem the hypotenuse satisfies c² = a² + b² = 36 + 64 = 100. Taking the positive square root gives c = 10 cm.
A logarithm gives the exponent to which the base must be raised to produce the argument. Since 2⁵ = 32, it follows that log₂ 32 = 5.
A circle centered at the origin with radius r has equation x² + y² = r². Substituting r = 3 gives x² + y² = 9.
By definition F(s) = ∫₀^∞ e^(−st)·e^(2t) dt = ∫₀^∞ e^(−(s−2)t) dt. This integral converges for s > 2 and evaluates to 1/(s − 2).
Integration by parts with u = x and dv = eˣ dx gives ∫ x·eˣ dx = x·eˣ − eˣ. Evaluating from 0 to 1 yields (1·e − e) − (0 − 1) = 0 + 1 = 1.
Squaring first gives z² = (1 + i)² = 1 + 2i + i² = 2i, since i² = −1. Then z⁴ = (z²)² = (2i)² = 4i² = −4.
The equation is quadratic in sin x and factors as (2 sin x + 1)(sin x − 1) = 0, so sin x = 1 or sin x = −1/2. The latter gives x = 7π/6 and 11π/6, both larger than π/2, so the smallest positive root is x = π/2.
Rewrite the expression as 3·[(sin 3x)/(3x)]. Since (sin u)/u tends to 1 as u tends to 0, the limit equals 3 × 1 = 3.
For an arithmetic series the sum of n terms is Sₙ = (n/2)(2a + (n − 1)d). Here a = 3, d = 4 and n = 20, so S₂₀ = 10(6 + 76) = 820.
A square root is defined only when its radicand is non-negative, so the condition is 2x − 6 ≥ 0. Solving gives x ≥ 3, and x = 3 itself is permitted because √0 = 0.
Composition works from the inside out, so evaluate g(2) = 4 first. Then f(4) = 2(4) + 1 = 9.
The sum of two logarithms with the same base equals the logarithm of the product, so log₁₀ 2 + log₁₀ 5 = log₁₀ 10. Because 10¹ = 10, the value is 1.
Arranging the values in order gives 3, 4, 4, 7, 8, 9. Because there is an even number of values, the median is the mean of the two middle entries: (4 + 7)/2 = 5.5.
Using the quotient identity cot θ = cos θ / sin θ, the product becomes sin θ · (cos θ / sin θ). The sine factors cancel, leaving cos θ.
Both squared terms appear with positive coefficients, and the two denominators are different. That is the standard form of an ellipse whose semi-axes are 3 and 4.
Write y = 3x − 7 and solve for x, giving x = (y + 7)/3. Interchanging x and y then produces f⁻¹(x) = (x + 7)/3.
Expanding along the first row gives 1·(1·0 − 4·6) − 2·(0·0 − 4·5) + 3·(0·6 − 1·5). That equals −24 + 40 − 15 = 1.
The general term is C(5, k)·2^(5−k)·x^k, so the x³ term uses k = 3. Its coefficient is C(5, 3)·2² = 10 · 4 = 40.
The word has 6 letters with A repeated 3 times and N repeated 2 times. Dividing the total permutations by the repeats gives 6!/(3!·2!) = 720/12 = 60.
The outcomes giving a sum of 8 are (2,6), (3,5), (4,4), (5,3) and (6,2), five equally likely ordered pairs. Exactly one of these has 3 on the first die, so the conditional probability is 1/5.
The mean is 6, so the squared deviations are 16, 4, 0, 4 and 16. Their sum is 40, and dividing by the 5 data points gives a variance of 8.
Dividing numerator and denominator by x² gives (3 + 2/x)/(1 − 5/x²). As x grows without bound the terms containing 1/x and 1/x² vanish, leaving 3/1 = 3.
Differentiating both sides with respect to x gives 2x + 2y(dy/dx) = 0. Solving for the derivative yields dy/dx = −x/y.
The slope of the tangent equals the derivative, and the derivative of sin x is cos x. At x = π/3 the cosine equals 1/2.
Let u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫e^u du, which is (1/2)e^(x²) + C.
The cross product is (1·0 − 0·3, 0·1 − 2·0, 2·3 − 1·1) = (0, 0, 5). Its magnitude is therefore 5.
Writing 5x + 1 = A(x + 2) + B(x − 1) and substituting x = 1 gives 6 = 3A, so A = 2. Substituting x = −2 gives −9 = −3B, so B = 3.
The complex number √3 + i has modulus 2 and argument 30°. De Moivre's theorem gives modulus 2⁶ = 64 and argument 6·30° = 180°, and 64 at 180° equals −64.
The curves meet where x² = 2x − x², giving x = 0 and x = 1, and 2x − x² is the upper curve on that interval. The area is the integral of 2x − 2x² from 0 to 1, which equals [x² − (2/3)x³] evaluated from 0 to 1, namely 1/3.
A p-series with terms 1/n^p converges exactly when p > 1. In the second option p = 2, while p = 1 and p = 1/2 both give divergent p-series.
An antiderivative of x^(−3) is −1/(2x²). Evaluating from 1 to infinity gives 0 − (−1/2) = 1/2, so the improper integral converges.
Setting f′(x) = 3x² − 3 = 0 gives the critical point x = 1, where f(1) = −2. Comparing this with the endpoint values f(0) = 0 and f(2) = 2 shows the maximum on the interval is 2.
The characteristic equation r² − 5r + 6 = 0 has roots r = 2 and r = 3, so y = C₁e^(2t) + C₂e^(3t). The conditions give C₁ + C₂ = 1 and 2C₁ + 3C₂ = 0, whose solution is C₁ = 3, C₂ = −2.