What is the solution set to 8(2x - 3) s 20

When 2/3 of a number is increased by 20, the sum is then halved, the result obtained is the same as 2/3 of the number, increased

aleksklad [387]

$\frac{2/3x+20}{2} =\frac{2}{3} x+3\\\frac{2/3x+20}{2} - 3 = \frac{2}{3}x\\\frac{2/3x+20}{2} - \frac{6}{2} = \frac{2}{3}x\\\frac{2/3x+20-6}{2} = \frac{2}{3}x\\\frac{2/3x+14}{2} = \frac{2}{3}x\\2(\frac{2/3x+14}{2}) = 2(\frac{2}{3}x)\\2/3x+14=\frac{4}{3}x\\14=\frac{4}{3}x-\frac{2}{3}x\\14=\frac{2}{3}x\\\frac{14}{1}/\frac{2}{3}=x\\\frac{14}{1}*\frac{3}{2}=x\\7*3=x\\x=21$

The number is 21.

4 0

3 years ago

Last year Victoria paid £354 for her car insurance

Fed [463]

Answer:

the price decreased by 7.06 percent

Step-by-step explanation:

100/354x329=92.93785311

100-92.93785311=7.06

3 0

3 years ago

Write the number that is five less than ten million.

andrey2020 [161]

Answer:

9,999,995

Step-by-step explanation:

8 0

3 years ago

Todd drives 60mph. How long will it take him to go 300 miles?

Paraphin [41]

300/60=5 so your answer is 5 hours

6 0

3 years ago

Read 2 more answers

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .

3 0

3 years ago

What is the solution set to 8(2x - 3) s 20 - 4X?