Answer:What are the equivalence classes of the equivalence relations in Exercise 3? A binary relation defined on a set S is said to be equivalence relation if it is reflexive, symmetric and transitive. An equivalence relation defined on a set S, partition the set into disjoint equivalence classes

7 0

3 years ago

Solve with interval notation Solve: 5 | x − 2 | + 4 > 8

Gnesinka [82]

Answer:

x<6/5, x>14/5

Step-by-step explanation:

Steps

$5\left|x-2\right|+4>8$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$5\left|x-2\right|+4-4>8-4$

$\mathrm{Simplify}$

$5\left|x-2\right|>4$

$\mathrm{Divide\:both\:sides\:by\:}5$

$\frac{5\left|x-2\right|}{5}>\frac{4}{5}$

$\mathrm{Simplify}$

$\left|x-2\right|>\frac{4}{5}$

$\mathrm{Apply\:absolute\:rule}:\quad\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:<\:-a\:\quad\mathrm{or}\quad\:u\:>\:a$

$x-2<-\frac{4}{5}\quad\mathrm{or}\quad\:x-2>\frac{4}{5}$

Show Steps

$x-2<-\frac{4}{5}\quad:\quad x<\frac{6}{5}$

Show Steps

$x-2>\frac{4}{5}\quad:\quad x>\frac{14}{5}$

$\mathrm{Combine\:the\:intervals}$

$x<\frac{6}{5}\quad\mathrm{or}\quad\:x>\frac{14}{5}$

5 0

2 years ago

Multiply using FOIL or Table method (x + 4) (x + 5)

mylen [45]

Answer:

${x}^{2} + 9x + 20$

Step-by-step explanation:

1) Use FOIL method: (a + b) (c + d) = ac + ad + bc + bd.

${x}^{2} + 5x + 4x + 20$

2) Collect like terms.

${x}^{2} + (5x + 4x) + 20$

3) Simplify.

${x}^{2} + 9x + 20$

Therefor, the answer is x² + 9x + 20.

5 0

2 years ago

Evaluate f(-2) if f(x) = x + 3 OC bh

natka813 [3]

Answer: is this 8th grade math or high school or college because i am in 8th grade

Step-by-step explanation:

8 0

2 years ago

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