Lily ran 1,000 meters in 12 minutes. How many meters/minute did she run?

Mathematics

1 answer:

vladimir2022 [97]3 years ago

5 0

Answer:

1,000 divided by 12=

aprox 83.3 meters per minute

You might be interested in

Hey can you please help me posted picture of question

larisa [96]

$(5x - 2)^2 = 15$

Square root both sides:
$5x - 2 = \pm\sqrt{15}$

Add 2 to both sides:
$5x = pm\sqrt{15} + 2$

Divide both sides by 5:
$x = \dfrac{ \pm\sqrt{15} + 2}{5}$

Answers:
$x = \dfrac{ \sqrt{15} + 2}{5} \ or \ \dfrac{ - \sqrt{15} + 2}{5}$

Answer: (B) and (E)

5 0

3 years ago

Read 2 more answers

Suzanne in Barry each by four equals sized bagels. They divide the bagels equally among themselves and three other friends. How

grandymaker [24]

Number of people = 5 (Suzanne, Barry, and 3 friends)
Number of bagels = 4

4 bagels divided between 5 people is 4/5.

6 0

3 years ago

Read 2 more answers

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .