All categories

2 years ago

Is god real???? Answer ASAP

Mathematics

2 answers:

marusya05 [52]2 years ago

8 0

depends on what u believe in if u think he existed then yes if you dont then no

OlgaM077 [116]2 years ago

6 0

Id like to think so but it also depends on what you believe like do you think that he created this whole universe or do you think something like that isn’t possible i believe he is because i know that everything happens for a reason and he knows all the reasons and has everything planned out

You might be interested in

X2 - 10xy + 3y + y2 - 1 2 3 4 5

lesantik [10]

Answer:

Answer: There are 5 (five terms) in the expression shown.

Step-by-step explanation:

_______________________________________________________

In the expression shown:

___________________________________________________

" x² − 10xy + 3y + y² − 1 " ;

___________________________________________________

→ These 5 (five terms) are:

______________________________________________

x² , -10xy, 3y, y², and -1 .

______________________________________________

5 0

3 years ago

Maria has baked a cake for Cathy's birthday. The dimensions of the cake are 13 1/2 inches by 6 3/4 inches. The cake is going to

ycow [4]

Answer:

Step-by-step explanation:

Given:

Length = 27/4 in

Width = 27/2 in

5.55% of the cake = 5.55/100

= 1/18

Therefore, 18 equal squares of cake where 1 piece is 5.55 % of the cake

Since, the cake is twice as long as it is wide, so that means cut half the cake (27/4 × 27/4) into 9 pieces.

Therefore, each has a side of 1/3 * 27/4

= 9/4 inches.

8 0

3 years ago

X=-12-6y <br>-4x+5y=-39<br>slove by the system of substitution

Artist 52 [7]

X = -6y - 12
4x + 5y =-39

4x + 5y = -39
-4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
24y + 48 + 5y = -39
29y + 48 = -39
29y = -87
y = -3

x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
x = 6

(x, y) = (6, -3)

5 0

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