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3 years ago

Craig had p coins. Then he found 66 more coins in a drawer. Write an expression that shows how many coins Craig has now.

1 answer:

5 0

p+ 66

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Whats the leastf 3/4, 4/5, and 2/3

timurjin [86]

Answer: 2/3

Step-by-step explanation: convert to same unit. 45/60, 48/60, 40/60

40/60<45/60<48/60

so 2/3 is least

4 0

2 years ago

If they are both right will give the thing and 5star if all are right btw

Vlad [161]

Answer:

The answer is

4. n ≤ −4

5. x ≤ −10

I hope this helps you

7 0

2 years ago

Read 2 more answers

Can someone help with trig identities?<br> (calculators are allowed)

Oksana_A [137]

Answer:

Step-by-step explanation:

cosA^2 = 1 - sinA^2

subtitute 1/4 below

= 1 - (1/4)^2

= 1 - 1/16

after calculation

= - 0.9682

8 0

2 years ago

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

3. Round off the following numbers to three decimal places:<br>a) 31,1104<br>b) 0,10989

Luba_88 [7]

A) 31.110
B) 0.110

When rounding to three decimal places, you consider if the number to the right of the third decimal place is less than or greater than 5. If it's less than 5, then the digit in the third decimal place stays the same. If it is greater than or equal to 5, then you round the digit in the third decimal place up by one.

4 0

2 years ago

Read 2 more answers