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3 years ago

Can someone help, i dont need step by step or anything <3

Mathematics

1 answer:

Leviafan [203]3 years ago

5 0

The answer is 145. I used the other angles as a triangle; I forgot what the theory is called for the easier way. I found out the complementary angle from subtracting <1 from 180. Then I added the new angle and <2 this gives me <4. It would be easier to explain if I remembered what the theory was called. Sorry and hope this helps; this is the answer bye.

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Compute the difference quotient, StartFraction f (x plus h )minus f (x )Over h EndFraction f(x+h)−f(x) h, for the function f (

lawyer [7]

Answer:

The difference quotient for $f(x)=3x^2$ is $3 h + 6 x$ .

Step-by-step explanation:

The difference quotient is a formula that computes the slope of the secant line through two points on the graph of f. These are the points with x-coordinates x and x + h. The difference quotient is used in the definition the derivative and it is given by

$\frac{f(x+h)-f(x)}{h}$

So, for the function $f(x)=3x^2$ the difference quotient is:

To find $f(x+h)$ , plug $x+h$ instead of $x$

$f\left(x+h\right)=3 \left(h + x\right)^{2}$

Finally,

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\left(3 \left(h + x\right)^{2}\right)-\left(3 x^{2}\right)}{h}$

$\frac{3\left(\left(h+x\right)^2-x^2\right)}{h} \\\\\frac{3(h^2+2hx+x^2-x^2)}{h} \\\\\frac{3h\left(h+2x\right)}{h}\\\\3h+6x$

The difference quotient for $f(x)=3x^2$ is $3 h + 6 x$ .

4 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 $3 \sim 2$ .