Given: KLMN is a trapezoid, KL=MN, m∠1=m∠2, LM/KN = 8/9 , Perimeter KLMN=132 Find: The length of midsegment.

Answer:

A =  84 inches^2

Step-by-step explanation:

We know that the formula for the area of a triangle is given by

A = 1/2 b*h

Let's substitute what we know

We need the units to be the same

Convert 7 ft to inches

1 ft = 12 inches

Multiply both sides by 7

7 ft = 84 inches

A = 1/2 *84*2

A =  84 inches^2

Answer:

The product of 4 and m is -8.

Without numbers: The product of four and the variable m is negative eight.

Step-by-step explanation:

4m means m is multiplied by 4. The result of the multiplication operation is called a "product." The equal sign translates to "is".

The sentence 'Four times a certain number equals negative eight' corresponds to the equation 4m = -8, indicating that multiplying a number by four yields negative eight.

The sentence to represent the equation 4 m = -8 might be: "Four times a certain number equals negative eight." This sentence encapsulates the equation by specifying that the product of the number m and four is equivalent to negative eight, implying that m will have a negative value since it is equal to a negative number when multiplied by a positive.

Answer:

The answer is the proof so it is long.

The question doesn't define u(n), but it's not hard to guess.

Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.

Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.

If n isn't prime, we exclude from the group all integers which share factors with n.

Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).

Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).

Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.

Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.

Answer:

The answer is the proof so it is long.

The question doesn't define u(n), but it's not hard to guess.

Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.

Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.

If n isn't prime, we exclude from the group all integers which share factors with n.

Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).

Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).

Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.

Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.