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I. Evaluate (8x7) (x) (5) 2(x) 6 12 ii. 3(X) P = 6 K(X)K = 4 iii. K(X)k = 4 K = 2

Mathematics

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Answers

3 parts
ii
1.4
a

P=2P=2

K=2 or K=−2K=2 \text{ or } K=-2

b
18

Step-by-step solution

9 steps

Here are the solutions to the problems from the image.

** (Evaluation):**

Part ii· 5 steps

Evaluate the expression based on the student's markings

The expression is given as (8(x)7)(x)(3/5)(8(x)7)(x)(3/5). Based on the numbers 2 and 12 written below the expression, and the (x) symbols likely representing multiplication, the expression is interpreted as:

8×72×12×35\frac{8 \times 7}{2 \times 12} \times \frac{3}{5}
  1. Multiply the numbers in the numerator and denominator of the first fraction.

    5624×35\frac{56}{24} \times \frac{3}{5}
  2. Simplify the first fraction.

    5624=7×83×8=73\frac{56}{24} = \frac{7 \times 8}{3 \times 8} = \frac{7}{3}
  3. Multiply the simplified fractions.

    73×35\frac{7}{3} \times \frac{3}{5}
  4. Cancel out the common factor of 3.

    7×33×5=75\frac{7 \times \cancel{3}}{\cancel{3} \times 5} = \frac{7}{5}
  5. Convert the fraction to a decimal.

    75=1.4\frac{7}{5} = 1.4

    The evaluated expression is 1.4\boxed{1.4}.

    ** (Equations):**

Part a

P=2P=2

Solve for P and K from the equations

3(X)P=63(X) P = 6 K(X)K=4K(X)K = 4
  1. Interpret (X) as multiplication.

    The equations become:

    3P=63P = 6 K×K=4  ⟹  K2=4K \times K = 4 \implies K^2 = 4
  2. Solve for P.

    3P=63P = 6 P=63P = \frac{6}{3} P=2P = 2
  3. Solve for K.

    K2=4K^2 = 4 K=±4K = \pm \sqrt{4} K=±2K = \pm 2

    The values are P=2\boxed{P=2} and K=2 or K=−2\boxed{K=2 \text{ or } K=-2}. (The student's work shows K=2K=2, implying a positive value).

Part b18

Solve for P from the equation

3()6=P3( )6 = P

This equation is ambiguous due to the empty parenthesis. Assuming it represents multiplication, the equation is:

3×6=P3 \times 6 = P
  1. Multiply the numbers.

    P=18P = 18

    The value of P is 18\boxed{18}.

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