Can someone help me please

Nikolay [14]

Both arrive after P Waves.

6 0

3 years ago

(2.3 x 105)(1.5 x 104) in scientific notation plssss ASAP

zhenek [66]

By using scientific notation we calculated (2.3 $\times$ 10⁵)(1.5 $\times$ 10⁴) we got 345 $\times$ 10⁷.

Given that,

(2.3 $\times$ 10⁵)(1.5 $\times$ 10⁴)

By using scientific notation

The scientific notation allows us to express extremely large or extremely small numbers by multiplying them by single-digit integers and by 10 raised to the power of the corresponding exponent. The exponent is either positive or negative depending on how big or little the number is.

Now, we calculate

(2.3 $\times$ 10⁵)=2.5 $\times$ 100000=230000

(1.5 $\times$ 10⁴)=1.5 $\times$ 10000=15000

Now,

=230000 $\times$ 15000

=3450000000

=345 $\times$ 10⁷

Therefore, By using scientific notation we calculated (2.3 $\times$ 10⁵)(1.5 $\times$ 10⁴) we got 345 $\times$ 10⁷.

To learn more about scientific notation visit: brainly.com/question/18073768

#SPJ1

3 0

1 year 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

I just need all the numbers on top in feet on the graph please

sattari [20]

Answer:

2.5, 5, 7.5

Step-by-step explanation:

Count.

3 0

2 years ago

A 32 foot tall tree casts a shadow that is 48 feet long. How far away from the tree is a

sveticcg [70]

Answer:

39 feet

Step-by-step explanation:

take the tree as AB , the distance between the man and the tree as BC, the distance between the man and the tip of the shadow CD and the point intersecting the hypotenuse CE

since CE parallel to AB we can BPT

CE/AB = CD/BD

6/32= CD/48

CD = 9 feet

since CD is feet

BC is 48-9 = 39 feet

3 0

1 year ago