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3 years ago

1.588 x 10 to the negative 4th power

Mathematics

2 answers:

pshichka [43]3 years ago

7 0

Answer:

1.588 * 10 = 15.88.
15.88 to the negative 4th power is 0.00001572526

damaskus [11]3 years ago

4 0

1588 is the answer hope this helps you

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HELP MEEEEEEEEEE Package A contains 3 birthday cards and 2 thank-you notes and costs $9.60. Package B contains 8 birthday cards

hodyreva [135]

Answer:

Each birthday card costs $2.2 ⇒ answer c

Step-by-step explanation:

* Lets explain how to solve the problem

- Package A contains 3 birthday cards and 2 thank-you notes

- It costs $9.60

- Package B contains 8 birthday cards and 6 thank-you notes

- It costs $26.60

- x represents the cost of birthday card and y represents the cost of

thank-you note

* Lets change these information to two equations

∵ x represents the cost of each birthday cards

∵ y represents the cost of each thank-you notes

∵ Bag A contains 3 birthday cards and 2 thank-you notes

∵ Bag A costs $9.60

∴ 3x + 2y = 9.60 ⇒ (1)

∵ Bag B contains 8 birthday cards and 6 thank-you notes

∵ Bag B costs $26.60

∴ 8x + 6y = 26.60 ⇒ (2)

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- Multiply equation (1) by -3 to eliminate y

∵ -3(3x) + -3(2y) = -3(9.60)

∴ -9x - 6y = -28.8 ⇒ (3)

- Add equations (2) and (3)

∴ -x = -2.2

- Multiply both sides by -1

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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.

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2 years ago