Will give brainleist

If b = 7, what is the value of the expression 2(10 – b)?

Stella [2.4K]

Answer:

6

Step-by-step explanation:

10-7=3

2*3=6

4 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!! DUE TODAY!!!!!

prisoha [69]

Answer:

12.4 cm³

Step-by-step explanation:

From the picture attached,

Radius of the circular top of the cone = $\frac{\text{Diameter}}{2}$ = $\frac{2}{2}=1$

Height of the cone = h

Lateral height of the cone = h

By applying Pythagoras theorem in the right triangle of the cone,

l² = r² + h²

6² = 1² + h²

h = $\sqrt{36-1}$

h = $\sqrt{35}$

Ice cream needed to fill one cone = Volume of the cone

Since, formula for the volume of the cone V = $\frac{1}{3} \pi r^{2} h$

V = $\frac{1}{3}\pi (1)^2(\sqrt{35})$

= 6.195

≈ 6.20 cm³

Ice cream needed to fill the two cones = 2 × 6.20

= 12.4 cm³

Therefore, ice cream needed to fill the two cone = 12.4 cm³

4 0

3 years ago

(HELP FINAL EXAM) If two states are selected at random from a group of 40 states, determine the number of possible outcomes if

Hitman42 [59]

Answer:

With replacement, there are 1600 possible outcomes.

Without replacement, there are 1560 possible outcomes.

Step-by-step explanation:

Chosen with replacement:

State is chosen, then put back in the group, then another state can be chosen. In both cases, there will be 40 possible outcomes, as the first taken state is placed back. So

40*40 = 1600

With replacement, there are 1600 possible outcomes.

Chosen without replacement:

State and chosen and is not replaced.

So, for the second state, there will be only 39 possible outcomes.

40*39 = 1560

Without replacement, there are 1560 possible outcomes.

7 0

2 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 $3 \sim 2$ .