What is the value of 3² + 2³?
- 13
- 36
- 17 Correct answer.
- 25
Explanation
Evaluate each power first: 3² = 9 and 2³ = 8. Adding gives 9 + 8 = 17, since exponentiation is applied before addition.
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Evaluate each power first: 3² = 9 and 2³ = 8. Adding gives 9 + 8 = 17, since exponentiation is applied before addition.
Subtract 3x from both sides to get 2x − 7 = 9, then add 7 to obtain 2x = 16. Dividing both sides by 2 gives x = 8.
The two given sides are the legs of a right triangle, so the third side is the hypotenuse. By the Pythagorean theorem, c² = 6² + 8² = 36 + 64 = 100, hence c = 10 cm.
A logarithm answers the question of what exponent of the base gives the argument. Since 2⁵ = 32, log₂ 32 = 5.
An irrational number cannot be written as a ratio of two integers. The square root of 2 is not a perfect square, so √2 cannot be expressed as a fraction and is irrational.
The interior angle sum of an n-gon is (n − 2)·180°. Setting (n − 2)·180 = 900 gives n − 2 = 5, so n = 7.
Substitute x = −2 carefully: 2(−2)² = 2·4 = 8, then −3(−2) = +6, then +1. Summing gives 8 + 6 + 1 = 15.
Slope equals the change in y divided by the change in x. Here (11 − 3)/(6 − 2) = 8/4 = 2.
For the standard acute angles, sin 30° = 1/2, coming from the half-equilateral triangle where the side opposite 30° is half the hypotenuse.
The discriminant is b² − 4ac = (−4)² − 4·2·5 = 16 − 40 = −24. A negative discriminant means the quadratic has no real roots, so its two roots form a complex conjugate pair.
The nth term of an arithmetic sequence is a₁ + (n − 1)d. Here 7 + 14·4 = 7 + 56 = 63.
Order does not matter in a committee, so we use combinations: C(8,3) = 8·7·6 / (3·2·1) = 56.
The outcomes greater than 4 are 5 and 6, giving 2 favorable outcomes out of 6 equally likely ones. The probability is 2/6 = 1/3.
When multiplying powers of the same base, add the exponents: x³·x⁵ = x⁸. When dividing, subtract the exponents: x⁸/x² = x⁶.
A point on the x-axis has y = 0. Substituting gives 2x = 12, so x = 6, and the intercept is (6, 0).
If the roots are r and s, then the sum is −b/a and the product is c/a. Here the sum is −2, so the x-coefficient is +2, and the product is −15, so the constant is −15, giving x² + 2x − 15 = 0.
For an infinite geometric series with first term a and ratio r where |r| < 1, the sum is a/(1 − r). Here a = 1 and r = 1/3, giving 1/(1 − 1/3) = 1/(2/3) = 3/2.
Using the 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.
The expression under an even root must be non-negative, so x − 3 ≥ 0. That gives x ≥ 3, including x = 3 where the root is zero.
Arc length equals rθ with θ in radians. Converting 72° gives 72·π/180 = 2π/5 radians, so the arc is 5·(2π/5) = 2π.
Areas of similar figures scale as the square of the linear ratio. Squaring 3 : 5 gives 9 : 25.
A quadratic has a double root exactly when its discriminant is zero. Here b² − 4ac = k² − 36 = 0, so k² = 36 and k = ±6.
The identity 1 + tan²θ = sec²θ holds for all defined θ. Multiplying by cos²θ gives sec²θ·cos²θ = 1, since sec θ = 1/cos θ.
In a geometric sequence aₙ = a₁rⁿ⁻¹, so a₅/a₂ = r³. Here 48/6 = 8 = r³, giving r = 2.
There are 6 letters with repeated letters: A appears 3 times and N appears 2 times. Dividing the total permutations by the factorials of the repeats gives 6!/(3!·2!) = 720/12 = 60.
For the standard angle 60°, the cosine equals 1/2, as seen in the 30-60-90 triangle where the side adjacent to 60° is half the hypotenuse.
The intersection A ∩ B contains exactly the elements that belong to both sets. The only element present in both A and B is 3.
For a 2×2 matrix [[a, b], [c, d]] the determinant equals ad − bc. Here that is 2·3 − 5·1 = 6 − 5 = 1.
The modulus of a + bi is √(a² + b²). Here √(3² + (−4)²) = √(9 + 16) = √25 = 5.
Differentiate term by term using the power rule: the derivative of x³ is 3x², and the derivative of −4x is −4. The constant term would vanish, but here none is present.
An antiderivative of 2x is x². Evaluating from 0 to 1 gives 1² − 0² = 1.
By definition of the logarithm, log₃ x = 2 means 3² = x. Therefore x = 9.
For a parabola y = ax² + bx + c, the vertex x-coordinate is −b/(2a). Here −(−6)/(2·1) = 3.
Since θ is acute with sin θ = 4/5, we get cos θ = 3/5 from the Pythagorean identity. Then sin 2θ = 2 sin θ cos θ = 2·(4/5)·(3/5) = 24/25.
We are arranging 4 distinct digits chosen from 5, which gives 5·4·3·2 = 120 ordered arrangements.
Two vectors are perpendicular exactly when their dot product is zero. Setting 2k + (−1)(3) = 0 gives 2k = 3, so k = 3/2.
The sum of the first n odd integers equals n². With n = 20, the sum is 20² = 400.
Dividing P(x) by (x − 1) gives x² − x − 6, which factors as (x − 3)(x + 2). The remaining roots are therefore 3 and −2.
A circle with center (h, k) and radius r has equation (x − h)² + (y − k)² = r². Substituting h = 2, k = −3, r = 4 gives (x − 2)² + (y + 3)² = 16.
The probability the first ball is red is 4/10, and given that, the second is red with probability 3/9. Multiplying gives (4/10)(3/9) = 12/90 = 2/15.
Factor the numerator as (x − 2)(x + 2) and cancel the common factor (x − 2), valid for x ≠ 2. The expression becomes x + 2, whose limit as x → 2 is 4.
Set y = 2x + 3 and solve for x: x = (y − 3)/2. Interchanging x and y gives f⁻¹(x) = (x − 3)/2.
C(6, 2) = 15 and C(6, 3) = 20, so the sum is 35. This also equals C(7, 3) by Pascal's identity.
By Vieta's formulas, the sum of the roots of ax² + bx + c = 0 is −b/a. Here −(−12)/3 = 4.
Expand (1 + i)² = 1 + 2i + i². Since i² = −1, this simplifies to 1 + 2i − 1 = 2i.
Integrating f'(x) = 3x² gives f(x) = x³ + C. The condition f(0) = 2 forces C = 2, so f(x) = x³ + 2 and f(1) = 3.
The term containing x³ is C(5, 3)·2²·x³ = 10·4·x³ = 40x³, so the coefficient is 40.
An antiderivative of 1/x is ln x. Evaluating from 1 to e gives ln e − ln 1 = 1 − 0 = 1.
Apply the recurrence step by step: a₂ = 2·1 + 1 = 3, a₃ = 2·3 + 1 = 7, a₄ = 2·7 + 1 = 15.
Infinitely many solutions occur when the two equations are proportional. The second equation is exactly twice the first when m = 2·3 = 6 and the constant 10 = 2·5, giving a dependent system.