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If 3245six = xnine, find x. Convert 101.01two to base ten. Solve (x+2)/3 = 5.

Mathematics

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2 parts
ii

5.375ten5.375_{ten}

iii
13

Step-by-step solution

5 steps

Problem (i) is ambiguous due to unclear notation. The expression "xninex_{nine}" is not standard. If it means the number xx in base nine, and xx is a digit, then xx must be less than 9. However, for 3245x3245_x to be a valid number, the base xx must be greater than the largest digit, which is 5. This implies x∈{6,7,8}x \in \{6, 7, 8\}. Setting 3245x=x3245_x = x leads to the equation 3x3+2x2+4x+5=x3x^3 + 2x^2 + 4x + 5 = x, which simplifies to 3x3+2x2+3x+5=03x^3 + 2x^2 + 3x + 5 = 0. This equation has no positive integer solutions. Without further clarification or correction of a typo, problem (i) cannot be solved.

Here are the solutions for problems (ii) and (iii):

Part ii· 3 steps

Convert 101.011two101.011_{two} to base ten.

  1. Expand the binary number using powers of 2.

    101.011two=1×22+0×21+1×20+0×2−1+1×2−2+1×2−3101.011_{two} = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3}

  2. Calculate the values of the powers and multiply.

    101.011two=1×4+0×2+1×1+0×12+1×14+1×18101.011_{two} = 1 \times 4 + 0 \times 2 + 1 \times 1 + 0 \times \frac{1}{2} + 1 \times \frac{1}{4} + 1 \times \frac{1}{8} 101.011two=4+0+1+0+14+18101.011_{two} = 4 + 0 + 1 + 0 + \frac{1}{4} + \frac{1}{8}

  3. Sum the values.

    101.011two=5+28+18101.011_{two} = 5 + \frac{2}{8} + \frac{1}{8} 101.011two=5+38101.011_{two} = 5 + \frac{3}{8} 101.011two=5.375ten101.011_{two} = 5.375_{ten}

    The converted number is 5.375ten\boxed{5.375_{ten}}.

Part iii13

Solve x+23=5\frac{x+2}{3} = 5.

  1. Multiply both sides of the equation by 3 to eliminate the denominator.

    x+23×3=5×3\frac{x+2}{3} \times 3 = 5 \times 3 x+2=15x+2 = 15

  2. Subtract 2 from both sides of the equation to isolate xx.

    x=15−2x = 15 - 2 x=13x = 13

    The solution for xx is 13\boxed{13}.

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