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3 years ago

If you have 8 slices of pis and gave away 3/4 how many are left

Mathematics

2 answers:

Maru [420]3 years ago

3 0

8 - 3/4 = 4.6 let me know if I made any mistakes =^-^=

WARRIOR [948]3 years ago

3 0

Here, you need to simply subtract 3/4 from 8. If you have a hard time working with fractions, then convert it into a decimal.

Fraction method:
8-3/4=7 1/4
Decimal method:
8-0.75=7.25

So your answer is 7.25 pieces or 7 1/4 pieces.

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Solve the system of equations.<br> 3x + y = 3 <br> x + y = 2

qwelly [4]

When solving system equations, we can use substitution method or elimination. Today I'm using substitution method.

First name the 2 equations.
3x + y = 3 (1)
x + y = 2 (2)

Now pick one equation and express one algebra in forms of the other.
From (2),
x = 2 - y (3)

Now substitute (3) into (1),
3(2-y) + y = 3
6 - 3y + y = 3
6 - 2y = 3
6 - 3 = 2y
y = 1.5

Now substitute y = 1.5 into (2)
x + 1. 5 = 2
x = 2 - 1.5
x = 0.5

Therefore the answer is x = 0.5 and y = 1.5

5 0

3 years ago

How much will you get paid if you work 28.5 hours for $7.75 per hour?

jonny [76]

$220.87 , (it would be $220.88 if you have to round up though

4 0

3 years ago

I need help (–

ExtremeBDS [4]

Answer:

Step-by-step explanation:

–

q+6)=

–

5q–30=

–

Add -5 to both sides

Subtract -5 from both sides

Multiply both sides by -5

Divide both sides by -5

Apply the distributive property

5q=

Add 30 to both sides

Divide both sides by 5 would be 69

6 0

3 years ago

8 is 80% of what number A. 10.0 B. 6.4 C. 0.1 D. 640.0

kaheart [24]

The correct answer is A. 10

8 divided by 0.8 = 10
10x0.8=8

3 0

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

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