Hi :) how to do try it 6?

To prove that N is a proper subset of M, show that every element of N is an element of M but there exists some elements of M which are not in N.

M= {4, 8, 12, 16, 20, 24...}

N= {12, 24, 36, 48, 60...}

Similar to example 6, 12 is a multiple of 4.

If x∈N,

then x= 12m for some positive integer m.

∴ x= 4(3m)

Hence, x is a multiple of 4.

i.e. x∈N.

Thus, every element of N is an element of M, but there exists some elements of M which are not elements of N. For example, 4, 8 and 16 are elements of M but are not elements of N. Thus, N is a proper subset of M. (shown)

Explanation:

∈ is the symbol of 'element of'. Thus, x∈N means that x is an element of N.

☆ Think of subsets like subtopics:

If differentiation is a subtopic of calculus, differentiation is a part of calculus but calculus is not only about differentiation since it includes integration too. Thus, there are numbers, which we call elements, in M that are not in N but all elements in N are in M.

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation: