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3 years ago

Can anyone help with this

Mathematics

2 answers:

vesna_86 [32]3 years ago

8 0

I would be able to but i left my maths book at school and that has the explanation in, sorry :(

Marat540 [252]3 years ago

4 0

(g +5)x(2g +g) when something is together like this,2g,then u multiply 2 x g.
but i'm not sure what the paper says . because i dont know what the letters mean.but here u go

A.(g+5)(2g+3)=(g+5)(2g+3)=(g)(2g)+(g)(3)+(5)(2g)+(5)(3)=2g2+3g+10g+15=2g2+13g+15

do you understand?

B.(h+1)(2h−3)=(h+1)(2h+−3)=(h)(2h)+(h)(−3)+(1)(2h)+(1)(−3)=2h2−3h+2h−3=2h2−h−3

C.(r−4)(2r−7)=(r+−4)(2r+−7)=(r)(2r)+(r)(−7)+(−4)(2r)+(−4)(−7)=2r2−7r−8r+28=2r2−15r+28

I hope it helps

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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 $3 \sim 2$ .

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3 years ago

If a seed is planted, it has a 80% chance of growing into a healthy plant.

AlladinOne [14]

Answer:

0.336

Step-by-step explanation:

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P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 8, r = 7, p = 0.8, and q = 0.2.

P = ₈C₇ (0.8)⁷ (0.2)⁸⁻⁷

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Hey there!