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3 years ago

X3 + 5y when x = 4, y = -3

Mathematics

2 answers:

Georgia [21]3 years ago

5 0

Answer:

Step-by-step explanation:

First, because of the order of operations, you would do 4^3 (or four to the power of three) first to get 64. 5×-3 is equal to -15, and a positive number plus a negative number is just that positive number minus a positive version of the negative number, if that even makes sense. Basically, 35 + -2 would be equal to 35-2. 64-15 is equal to 49, which is your answer!

I hope this helped you!

Alisiya [41]3 years ago

4 0

Answer:

Step-by-step explanation:

4^3=64

5(-3)=-15

64+(-15)=49

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fgiga [73]

Answer:

a b and d

Step-by-step explanation:

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3 years ago

4. Isabelle takes the bus to work. The bus ride to work costs her $2.50 each time. She could buy a bus pass for a one-time fee o

iren [92.7K]

Answer:

6 times.

Exclamation:

You can do 15 divided by 2,50 to get you answer, when you do that you get 6.

(The upcoming and this sentence is pre-written and copy and pasted in every brainly question or comment I write or answer.) If this answer helped you please consider giving it brainliest. If my answer was wrong and you got marked wrong for it, I deeply apologize and hope you will forgive me, since everyone makes mistakes sometimes. If you need me to elaborate on my answer or give further explanation on it, please ask and I will do so. If you need to explain your reasoning on your work feel free to use my words- word for word- without crediting me, the answer was made for you anyways! Hope yall learn from my answer and it helps you in the future with assignments, quizzes, test’s, and more!

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8 0

3 years ago

From the set (76, 77, 78), use substitution to determine which value of x makes the equation true

jeyben [28]

Answer:

The answer is C.

Step-by-step explanation:

Multiply 4 by 76, 77, and 78.

4 * 76 = 304

4 * 77 = 308

4 * 78 = 312

304 = 304, so C is the correct answer.

Hope this helps!

7 0

3 years ago

Evaluate this piecewise function at x = -8 and x = 4.

oee [108]

Answer:

option D :

f(-8) = -2 ; f(4) = 3

Step-by-step explanation:

f(-8) = ³√-8 = -2

f(4) = 3 (the constant function)

6 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.

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2 years ago