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2 years ago

Help please thank you

Mathematics

2 answers:

Maslowich2 years ago

8 0

Answer:

it does not say anthing it says its restikted

Step-by-step explanation:

Tema [17]2 years ago

3 0

Answer:

A: x = -4

Step-by-step explanation:

Remove Parentheses:

8x - 3 = 10x + 5

Move terms

8x - 10x - 3 = 5

8x - 10x = 5 + 3

Collect Like Terms

-2x = 5 + 3

-2x = 8

Divide

-2x = 8

x = -4

Answer

x = - 4

Hope this helped

Brainliest is appreciated.

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Nick cut a circular cookie into 5 equal slices. what is the angle measure of each slice?

NemiM [27]

Answer:

72°

Step-by-step explanation:

The cookie has a shape of a circle. A circle is the locus of a point such that its distance from a point (center) is always the same. The sum of angles around the circle center is 360°.

Let x be the measure of the angle in each slice. Therefore the sum of the total 5 slice is:

x + x + x + x + x = 360

5x = 360

Dividing through by 5:

5x / 5 = 360 / 5

x = 72°

Each slice has an angle measure of 72°

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2 years ago

Hiiiiooo❤️Can someone please help me out with my question

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The answer is D. 50 because x equals 7 so if you fill that in for 4x-3 and multiply it by 2, you get 50

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3 years ago

Read 2 more answers

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