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MariettaO [177]
3 years ago
15

A line is graphed in the coordinate plane and two slope triangles are shown

Mathematics
1 answer:
aniked [119]3 years ago
3 0
If you could post the picture I can help
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22222222x22222222222222
UNO [17]

Answer:

4.93827 x 10^19

Step-by-step explanation:

hope it helped

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3 years ago
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A population of caribou in a forest grows at a rate of 5% every year. If there are currently 248 caribou, which function represe
ANEK [815]

Answer: I think that the answer is C because 5% is 5 out of 100 and 5 out of 100 as a decimal is 0.05 and you are trying to find how much the population will have increased so you have to multiply 248 by 5% or you have to multiply 248 by 0.05

Step-by-step explanation:

6 0
3 years ago
What should you pay attention to when using credit?
Oxana [17]

Answer:

All of the Above.

Step-by-step explanation:

Monthly Payments is similar to length of time. Interest rate is the amount you pay back on a monthly scale.

4 0
2 years ago
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The area of the triangle below is 8.91 square inches. What is the length of the base?<br> 2.2 in
Inessa05 [86]

Answer:

8.1

Step-by-step explanation:

Area of a triangle = ½ × base × height = Area

Rearranged =

area/1/2xheight

The base is the length.

8.91/1/2x2.2

=8.1 inches

4 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).

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
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