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3 years ago

Solve for all values of x by factoring. X^2-x+1=3

Mathematics

2 answers:

balandron [24]3 years ago

7 0

Answer:

x = 2, x = -1

Step-by-step explanation:

AlladinOne [14]3 years ago

3 0

Answer:

x = 2, -1

Step-by-step explanation:

x² - x + 1 = 3

- 3 -3

x² - x - 2 = 0

(x -2)(x + 1)

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The newly elected president needs to decide the remaining 8 spots available in the cabinet he/she is appointing. If there are 14

Elanso [62]

Answer:

The members of the cabinet can be appointed in 121,080,960 different ways.

Step-by-step explanation:

The rank is important(matters), which means that the order in which the candidates are chosen is important. That is, if we exchange the position of two candidates, it is a new outcome. So we use the permutations formula to solve this quesiton.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

If there are 14 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?

Permutations of 8 from a set of 14. So

$P_{(14,8)} = \frac{14!}{(14-8)!} = 121,080,960$

The members of the cabinet can be appointed in 121,080,960 different ways.

3 0

3 years ago

Find the reference angle of <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20-%20%5Cpi%7D%7B7%7D%20" id="TexFormula1" title

Semenov [28]

Answer:

π7

Step-by-step explanation:

since is in the first quadrant, the reference angle is π7

3 0

3 years ago

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

Read each sentence. Then, use the drop-down menus to identify the types of figurative language used. I must be popular because m

zepelin [54]

Answer:

i must be popular because my phone is blowing up- hyperbole

her smile lit up my day-metaphor

the dog continued to bark, speaking with authority-personification

he wept like a faucet-simile

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Use a common denominator to write an equivelent fractions for each fraction.

sergejj [24]

Sorry for not giving you the calculation here -

5/12 and 2/9 have denominators divisible by only 3 right? Well The common denominator that can then change is 3 because that's the only number they have in common (except 1)

8 0

3 years ago