answer:1. **Identify the necessary conditions for constructing a triangle with sides (a, b, c):** [ a + b > c, quad a + c > b, quad text{and} quad b + c > a ] These inequalities must be satisfied to form a triangle with sides of lengths (a, b, text{and} c). 2. **Define the new lengths (a', b', text{and} c'):** [ a' = frac{a}{a+1}, quad b' = frac{b}{b+1}, quad c' = frac{c}{c+1} ] 3. **Express the necessary conditions for constructing a triangle with sides (a', b', c'):** [ a' + b' > c', quad a' + c' > b', quad text{and} quad b' + c' > a' ] 4. **Verify one inequality (because of symmetry, the proof for any one condition suffices):** [ a' + b' > c' ] Substitute the expressions for (a', b'), and (c'): [ frac{a}{a+1} + frac{b}{b+1} > frac{c}{c+1} ] 5. **Clear the denominators by multiplying throughout by ((a+1)(b+1)(c+1)):** [ left( frac{a}{a+1} + frac{b}{b+1} right) (a+1)(b+1)(c+1) > frac{c}{c+1}(a+1)(b+1)(c+1) ] 6. **Simplify the left-hand side:** [ a(b+1)(c+1) + b(a+1)(c+1) > c(a+1)(b+1) ] 7. **Distribute and combine like terms:** [ a(bc + b + c + 1) + b(ac + a + c + 1) > c(ab + a + b + 1) ] [ ab(c+1) + a(b+1) + bc + b + a(b+1) + bc + b > ab + ac + bc + c ] [ abc + ab + ac + a + abc + ab + bc + b > abc + ab + ac + ac + bc + c ] 8. **Simplify further by collecting all terms:** [ abc + 2ab + ac + a + bc + b > abc + ab + ac + bc + c ] [ abc + 2ab + a + b > c ] 9. **Observe that (2ab + a + b > c) holds given (a + b > c) and (ab > 0):** Since (2ab > 0), it confirms: [ abc + 2ab + a + b > c ] 10. **Conclude that if (a + b > c), then (frac{a}{a+1} + frac{b}{b+1} > frac{c}{c+1}):** Hence, if the inequalities for (a, b, c) are satisfied, the inequalities for (a', b', c') are also satisfied. [ boxed{text{If a triangle can be formed with sides } a, b, text{and} c, text{then a triangle can also be formed with sides } a', b', text{and } c'.} ] 11. **Counterexample for the converse statement:** Consider (a' = frac{3}{4}, b' = frac{3}{4}, c' = frac{9}{10}): - These values satisfy (a' + b' = frac{3}{4} + frac{3}{4} = frac{6}{4} > frac{9}{10}), and similar checks show the inequality conditions are met. However, solving for corresponding (a, b, text{and} c): [ a = 3, quad b = 3, quad c = 9 ] - Here, (a + b = 3 + 3 = 6 < 9), which breaks the triangle inequality condition. Thus, it is a counterexample showing: [ boxed{text{The converse of the statement is not true.}} ] (blacksquare)

answer:First, factor out the common term: [ (3x+4)(x-5) + (3x+4)(x-7) = (3x+4)((x-5) + (x-7)). ] Simplify the expression inside the parentheses: [ (x-5) + (x-7) = 2x - 12. ] Thus, the equation becomes: [ (3x+4)(2x-12) = 0. ] This can be further simplified to: [ 2(3x+4)(x-6) = 0. ] The roots are x = -frac{4}{3} from 3x+4=0 and x = 6 from x-6=0. The sum of the roots is: [ -frac{4}{3} + 6 = frac{-4 + 18}{3} = frac{14}{3}. ] Thus, the sum of all roots is boxed{frac{14}{3}}.

answer:To factorize the expression m^{2}-4, we recognize it as a difference of squares. The difference of squares formula is a^2 - b^2 = (a+b)(a-b). Applying this formula to our expression, where a = m and b = 2, we get: [ m^{2}-4 = (m)^2 - (2)^2 = (m + 2)(m - 2) ] Therefore, the factorization of m^{2}-4 is: [ boxed{(m+2)(m-2)} ]

answer:To find the probability of not getting an "e" after spinning the spinner two times, we first need to find the probability of not getting an "e" on a single spin. Since there are 4 sections of equal size, the probability of landing on any one section is 1/4. Therefore, the probability of not landing on "e" in a single spin is 1 - P(landing on "e") = 1 - 1/4 = 3/4. Now, we need to find the probability of not getting an "e" in two consecutive spins. Since the spins are independent events, we can multiply the probabilities of each individual event to find the combined probability. So, the probability of not getting an "e" on the first spin AND not getting an "e" on the second spin is: (3/4) * (3/4) = 9/16 Therefore, the probability of not getting an "e" after spinning the spinner two times is boxed{9/16} .