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QUESTION 4: Simplify fully. 4.1 a³ x a² 4.2 (x³ + 1)⁰ 4.3 x/x⁵ 4.4 2x²yx³y / 12x⁶y³

Mathematics

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Step-by-step solution

8 steps

Part 3

Part 3.1

How many cans will there be in figure: e) f)

  • e) (Figure b): This figure has 3 cans in the bottom row, 2 in the middle, and 1 on top. Total cans = 3+2+1=63 + 2 + 1 = 6.
  • f) (Figure c): This figure has 5 cans in the bottom row, 4 in the next, 3 in the next, 2 in the next, and 1 on top. Total cans = 5+4+3+2+1=155 + 4 + 3 + 2 + 1 = 15.
  • e) 6\boxed{6}
  • f) 15\boxed{15}

Part 3.2

How many rows are there in: a) b) c)

  • a) (Figure a): This figure has 1 can, which forms 1 row.
  • b) (Figure b): This figure has 3 cans in the bottom row, and 3 rows in total.
  • c) (Figure c): This figure has 5 cans in the bottom row, and 5 rows in total.
  • a) 1\boxed{1}
  • b) 3\boxed{3}
  • c) 5\boxed{5}

Part 3.3

How many cans will there be in the nth row?

Observing the pattern from the top row downwards:
The 1st row (from the top) has 1 can.
The 2nd row (from the top) has 2 cans.
The 3rd row (from the top) has 3 cans.
Following this pattern, the nth row (from the top) will have n cans.

  • n\boxed{n}

Part 3.4

How does the number of rows relate to the number of cans in the base (bottom) row?

From the figures:

  • Figure a: 1 can in the base, 1 row.
  • Figure b: 3 cans in the base, 3 rows.
  • Figure c: 5 cans in the base, 5 rows. The number of rows is equal to the number of cans in the base (bottom) row.
  • The number of rows is equal to the number of cans in the base (bottom) row.

Part 4

Simplify fully.

Part 4.1· 1 step

a3×a2a^3 \times a^2

  1. Apply the exponent rule xm×xn=xm+nx^m \times x^n = x^{m+n}.

    a3×a2=a3+2a^3 \times a^2 = a^{3+2} a3+2=a5a^{3+2} = a^5
    • a5\boxed{a^5}

Part 4.2· 1 step

(x3+1)0(x^3 + 1)^0

  1. Apply the exponent rule that any non-zero base raised to the power of 0 is 1.

    (x3+1)0=1(x^3 + 1)^0 = 1
    • 1\boxed{1}

Part 4.3· 2 steps

x3x5\frac{x^3}{x^5}

  1. Apply the exponent rule xmxn=xm−n\frac{x^m}{x^n} = x^{m-n}.

    x3x5=x3−5\frac{x^3}{x^5} = x^{3-5} x3−5=x−2x^{3-5} = x^{-2}
  2. Rewrite with a positive exponent using x−n=1xnx^{-n} = \frac{1}{x^n}.

    x−2=1x2x^{-2} = \frac{1}{x^2}
    • 1x2\boxed{\frac{1}{x^2}}

Part 4.4· 4 steps

2x3y×3x2y12x6y3\frac{2x^3y \times 3x^2y}{12x^6y^3}

  1. Multiply the terms in the numerator.

    2x3y×3x2y=(2×3)×(x3×x2)×(y×y)2x^3y \times 3x^2y = (2 \times 3) \times (x^3 \times x^2) \times (y \times y) =6x3+2y1+1= 6x^{3+2}y^{1+1} =6x5y2= 6x^5y^2
  2. Substitute the simplified numerator back into the expression.

    6x5y212x6y3\frac{6x^5y^2}{12x^6y^3}
  3. Simplify the coefficients and variables separately.

    For coefficients: 612=12\frac{6}{12} = \frac{1}{2}
    For xx terms: x5x6=x5−6=x−1=1x\frac{x^5}{x^6} = x^{5-6} = x^{-1} = \frac{1}{x}
    For yy terms: y2y3=y2−3=y−1=1y\frac{y^2}{y^3} = y^{2-3} = y^{-1} = \frac{1}{y}

  4. Combine the simplified terms.

    12×1x×1y=12xy\frac{1}{2} \times \frac{1}{x} \times \frac{1}{y} = \frac{1}{2xy}
    • 12xy\boxed{\frac{1}{2xy}}
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