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The least number divisible by both 6 and 10 is their lowest common multiple. Since 6 = 2 × 3 and 10 = 2 × 5, the LCM is 2 × 3 × 5 = 30. Every common multiple of 6 and 10 must be a multiple of 30, so 30 is the least possible positive value.
A 20% reduction multiplies the price by 0.80, and a further 10% reduction multiplies the already reduced price by 0.90. The combined multiplier is 0.80 × 0.90 = 0.72, which corresponds to a 28% reduction. The two percentages cannot simply be added because the second reduction applies to a smaller amount.
In the ratio 5:2, each part is 350 ÷ 5 = 70 g. The sugar amount is 2 parts, so it is 2 × 70 = 140 g. Both quantities must be scaled by the same factor to keep the ratio unchanged.
In standard form a number is written as A × 10^n, where 1 ≤ A < 10. Moving the decimal point 11 places to the right gives 7.2, so the exponent is −11. Therefore 0.000000000072 = 7.2 × 10^-11.
Simplify each surd by extracting square factors: √50 = √(25 × 2) = 5√2 and √18 = √(9 × 2) = 3√2. Adding like surds gives 5√2 + 3√2 = 8√2. Only surds with the same radical part can be added directly.
Apply the power to every factor: (2x^3 y^-2)^3 = 8x^9 y^-6. Dividing by 4x y^2 means subtracting exponents, giving 2x^8 y^-8, which is 2x^8 / y^8. Negative exponents move the factor to the denominator.
The next simultaneous flash occurs after the least common multiple of 12 and 18. Since 12 = 2^2 × 3 and 18 = 2 × 3^2, the LCM is 2^2 × 3^2 = 36 seconds. Common multiples are multiples of the LCM, and 36 is the first positive one.
Let x = 0.171717... Then 100x = 17.171717..., because multiplying by 100 shifts the repeating block by two places. Subtracting gives 99x = 17, so x = 17/99. The fraction is already in lowest terms because 17 is prime and does not divide 99.
For a measurement to 1 decimal place, the true value lies within 0.05 cm of the stated value. The upper bounds are 8.45 cm and 3.75 cm, so the upper bound of the area is 8.45 × 3.75 = 31.6875 cm². The upper bound is found by multiplying the upper bounds of both measurements.
For inverse proportion, the product xy is constant. Here k = 4 × 15 = 60, so y = 60/x. When x = 10, y = 60/10 = 6.
Squaring -4 gives 16. The product 3 × (-5) is -15, so subtracting -15 is the same as adding 15. Thus 16 + 15 = 31.
Since 1/16 = 0.0625, multiplying by 3 gives 0.1875. Equivalently, 3 ÷ 16 = 0.1875 exactly.
12.5% is the same as 1/8. Dividing 480 by 8 gives 60, so 12.5% of 480 is 60.
Dividing 504 by 8 gives 63, and 63 = 3^2 × 7. Therefore 504 = 2^3 × 3^2 × 7.
Multiply the coefficients: 3.2 × 4.5 = 14.4. Add the powers of 10: 10^5 × 10^-2 = 10^3. Then 14.4 × 10^3 = 1.44 × 10^4 in standard form.
Multiply numerator and denominator by √5 + 1. The denominator becomes (√5)^2 - 1^2 = 4. Then 6(√5 + 1)/4 simplifies to 3(√5 + 1)/2.
Write y = kx^2. Substituting gives 45 = k × 3^2 = 9k, so k = 5. When x = 5, y = 5 × 5^2 = 125.
For a lower bound of a quotient, use the lower bound of the numerator and the upper bound of the denominator. With 1 d.p. measurements, a_lower = 12.25 and b_upper = 4.75. Then 12.25 ÷ 4.75 ≈ 2.5789, which rounds to 2.58.
The exponent 2/3 means cube root then square. The cube root of 27/8 is 3/2, and squaring gives 9/4. The negative sign takes the reciprocal, so the result is 4/9.
For N = p^a q^b r^c, the number of positive factors is (a+1)(b+1)(c+1). Here that is (4+1)(2+1)(1+1) = 5 × 3 × 2 = 30.
Leading zeros before the first non-zero digit are not significant. The first two significant digits are 4 and 8, and the next digit 7 rounds the 8 up to 9, so the value is 0.0049.
Writing 84 = 2^2 × 3 × 7 and 120 = 2^3 × 3 × 5 shows the shared prime powers are 2^2 and 3. Their product is 12, so the highest common factor is 12.
VAT is 20% of the pre-VAT price, and 20% of £240 is £48. Adding this to £240 gives £288.
Each year the amount is multiplied by 1.03, so after two years it is 1500 × 1.03^2. This equals 1500 × 1.0609 = £1591.35.
A negative exponent means take the reciprocal of the base before applying the power. Thus (2/5)^(-2) = (5/2)^2 = 25/4.
Expanding gives √3 × √3 - 5√3 + 2√3 - 10. Since √3 × √3 = 3, the integer part is 3 - 10 = -7 and the surd part is -5√3 + 2√3 = -3√3, so the result is -7 - 3√3.
Multiplying the map length by the scale gives 6.4 × 25,000 = 160,000 cm. Since 100,000 cm = 1 km, this is 1.6 km.
A value rounded to 7.3 to one decimal place lies within 0.05 of 7.3. The lower bound 7.25 is included because it rounds up to 7.3, while 7.35 is excluded because it rounds to 7.4.
Write 27 as 3^3, so the equation is 3^(2x + 1) = 3^(3x - 3). Equating exponents gives 2x + 1 = 3x - 3, hence x = 4.
For Nk to be a perfect cube, every prime exponent must be a multiple of 3. The exponents 2 on 3 and 5 each need one more factor, so k = 3 × 5 = 15.
Substituting x = 4 into f(x) = 3x − 5 gives 12 − 5 = 7, because function notation asks for the output at that input. The value 12 is only the product 3 × 4, before the subtraction is applied.
A rational function is undefined where its denominator equals zero, and x − 2 = 0 when x = 2. Therefore every real number except 2 may be used as an input.
Subtract 3 from both sides to get 2x < 8, then divide by the positive number 2, which keeps the inequality direction unchanged. This gives x < 4.
f(g(x)) means substitute g(x) into f: (2x − 3)^2 + 1. Expanding gives 4x^2 − 12x + 9 + 1 = 4x^2 − 12x + 10.
Set y = (2x − 1)/3 and solve for x: 3y = 2x − 1, so 2x = 3y + 1 and x = (3y + 1)/2. Replacing y by x gives f^{-1}(x) = (3x + 1)/2.
Complete the square: x^2 − 6x + 5 = (x − 3)^2 − 4. The vertex form (x − h)^2 + k has vertex (h, k), so here the vertex is (3, −4).
Since 27 = 3^3, the equation becomes 3^(x + 1) = 3^3. Equating exponents gives x + 1 = 3, so x = 2.
A right shift by 3 replaces x with x − 3 inside the function, while an upward shift by 2 adds 2 outside. Combining these gives y = f(x − 3) + 2.
The critical values are x = 1, where the numerator is zero, and x = −2, where the denominator is zero and the expression is undefined. Testing intervals shows the quotient is positive for x < −2, negative for −2 < x < 1, and positive for x > 1; including x = 1 gives x < −2 or x ≥ 1.
The composite is f(g(x)) = √(x^2 − 1), so the expression under the square root must satisfy x^2 − 1 ≥ 0. This is equivalent to x^2 ≥ 1, whose solution is x ≤ −1 or x ≥ 1.
For x = -3, the condition x < 0 applies, so f(-3) = 2(-3) + 1 = -5. For x = 2, the condition x ≥ 0 applies, so f(2) = 2^2 = 4. Therefore f(-3) + f(2) = -5 + 4 = -1.
Because (x - 2)^2 is never negative, -(x - 2)^2 is never positive. Thus f(x) ≤ 5 for every real x, and the maximum value 5 occurs at x = 2. The range is (-∞, 5].
The equation |x - 3| = 4 splits into x - 3 = 4 or x - 3 = -4. Solving gives x = 7 or x = -1. Both values satisfy the original equation.
Factor x^2 - 4x - 5 as (x - 5)(x + 1). The product is negative between the roots -1 and 5 because the parabola opens upward. Therefore the solution is -1 < x < 5.
Rewrite the logarithmic equation in exponential form: x - 1 = 2^3 = 8. Adding 1 gives x = 9. This value also keeps the argument x - 1 positive.
The average rate of change on [a, b] is (f(b) - f(a))/(b - a). Here f(4) = 24 and f(1) = 3, so the rate is (24 - 3)/(4 - 1) = 21/3 = 7.
By the remainder theorem, dividing p(x) by x - 1 gives remainder p(1). Compute p(1) = 1 - 2 + k + 6 = k + 5. Setting k + 5 = 4 gives k = -1.
Replacing x by 2x compresses the graph horizontally by a factor of 1/2, since each x-value is reached at half the input. Subtracting 1 outside the function shifts the graph down 1 unit. Thus the description is a horizontal compression by 1/2 and a shift down 1.
Take natural logarithms of both sides: (x + 1)ln2 = (x - 1)ln5. Expanding and collecting x terms gives x(ln2 - ln5) = -ln5 - ln2, so x = (ln5 + ln2)/(ln5 - ln2). Since ln5 + ln2 = ln10 and ln5 - ln2 = ln(5/2), this equals ln10/ln(5/2).
To find the inverse, swap x and y in y = (x + 1)/(x - 2) and solve for y. This yields y = (2x + 1)/(x - 1), so f^(-1)(x) = (2x + 1)/(x - 1). The original function has horizontal asymptote y = 1, which is excluded from its range, so the inverse domain is x ≠ 1.