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3 years ago

Solve the equation for 0 ≤ x 2π 4sin^2x+1=-4sinx

Mathematics

1 answer:

finlep [7]3 years ago

5 0

Answer:

7pi/6, 11pi/6

Step-by-step explanation:

WRONG! edge 2021, its A, not BBBBBBB

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Convert the unit measure

Nimfa-mama [501]

41lbs is equal to 656oz

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3 years ago

The balance in Beth's checking account is $75. She must deposit enough money in her account to be able to pay the electric bill,

Evgen [1.6K]

75+d=110 subtract 75 from both sides

d=$35

So she needs to deposit at least $35 to be able to pay the electric bill.

5 0

3 years ago

Read 2 more answers

Anyone know the answer to this question?

denpristay [2]

Answer:

sorry i dont know how to solve this

Step-by-step explanation:

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4 years ago

The formula y - y1 = m(x - x1) is the point-slope form of the equation of a line where m is the slope of the line and (x, y) and

a_sh-v [17]

Answer:

The slope of a line that includes the points (4, -2) and (5, 0) is 2

Step-by-step explanation:

You know that the formula y - y1 = m(x - x1) is the point-slope form of the equation of a line where m is the slope of a line.

The line must include the points (4, -2) and (5, 0). So, being:

(x,y)= (4,-2)
(x1,y1)= (5, 0)

and replacing in the point-slope form of the equation of a line:

-2-0=m(4-5)

You solve the equation for m and get:

-2=m*(-1)

$m=\frac{-2}{-1}$

m=2

<u><em>The slope of a line that includes the points (4, -2) and (5, 0) is 2</em></u>

5 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 $3 \sim 2$ .

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3 years ago