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3 years ago

The graph shows motion of a train. Enter a value to complete the statement

Mathematics

2 answers:

olya-2409 [2.1K]3 years ago

8 0

Answer:

25, i had the same question

schepotkina [342]3 years ago

4 0

Answer:25

Cause it is I tried it

Step-by-step explanation:

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Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago

The liter is the basic unit of weight in the metirc system?

Ivanshal [37]

False. The liter is the basic unit of volume in the metric system. The basic unit of weight is the kilogram.

8 0

3 years ago

What is the value of a in the equation 3 a plus b equals 54, when b equals 9?

NikAS [45]

Answer:

a = 15

Step-by-step explanation:

Given

3a + b = 54 ← substitute b = 9

3a + 9 = 54 ( subtract 9 from both sides )

3a = 45 ( divide both sides by 3 )

a = 15

5 0

3 years ago

Read 2 more answers

The farmer had 28 cows. He wanted to bring them into the milking shed in groups of 7. How many groups of cows did the farmer bri

Ostrovityanka [42]

Answer:

(C) Subtract 7 from 28. Keep subtracting 7 from answer until he gets to 0. Then count the number of times 7 was subtracted.

This method is another suitable solution to problem

Step-by-step explanation:

Total number of cows the farmer has = 28 cows

Number of cows in 1 shed = 7

The number of sheds used to keep 28 cows is 28 ÷ 7 = 4 sheds

⇒4 sheds are needed to keep 28 cows.

Now choosing an alternate method:

1) Add 28 and 7 = 28 + 7 = 35 sheds

But 35 sheds ≠ 4 sheds

Hence, NOT AN APPROPRIATE METHOD.

2)Add 7 to 28. Keep adding 7 to the answer for a total of seven times.

28 + 7 .... 7 times = 28 + 7 + 7 + 7 + 7 /+ 7 + 7 + 7 = 77 ≠ 4 sheds

NOT AN APPROPRIATE METHOD.

3)Subtract 7 from 28. Keep subtracting 7 from answer until he gets to 0.

28 - 7 = 21

21 - 7 = 14

14 -7 = 7 and 7 - 7 - 0

Hence, in 4 times the answer is 0

So, the number of sheds required is 4

AN APPROPRIATE METHOD.

4) Multiply 28 by 7.

28 x 7 = 196, 196 ≠ 4 sheds

NOT A APPROPRIATE METHOD.

Hence, the ONLY alternate method to count the totals sheds required is

Subtract 7 from 28. Keep subtracting 7 from the answer until he gets to 0.

6 0

2 years ago

9000 dollars is placed in an account with an annual interest rate of 8%. How much will be in the account after 17 years, to the

Fed [463]

Answer:

$33,300.16

Step-by-step explanation:

7 0

3 years ago