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3 years ago

what is the inverse of the conditional statement? If a polygon the has five angles, then it is a Pentagon

Mathematics

2 answers:

den301095 [7]3 years ago

9 0

Answer: The inverse f the given statement is

"If a polygon is not a Pentagon, then it does not have five sides".

Step-by-step explanation: We are given to write the inverse of the following conditional statement:

"If a polygon the has five angles, then it is a Pentagon".

<u>INVERSE of a conditional statement :</u>

If the conditional statement is

"If p, then q",

then its inverse statement is given by

"If not q, then not p".

Therefore, the inverse of the given statement is

"If a polygon is not a Pentagon, then it does not have five sides".

cupoosta [38]3 years ago

9 0

Mathwords: Inverse of a Conditional. Negating both the hypothesis and conclusion of a conditional statement. :)

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Mark got 54 out of 66 in his maths test how many did he get wrong. Give your answer in its simplest form

alex41 [277]

Answer:

9/11 I think!

Step-by-step explanation:

54/66= 9/11

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2 years ago

The equation V = 16300 (0.94)^t represents the value (in dollars) of a car t years after its purchase. Use this equation to comp

Dafna1 [17]

Solution:

Given:

$V=16300(0.94)^t$

The value of a car after t - years will depreciate.

Hence, the equation given represents the value after depreciation over t-years.

To get the rate, we compare the equation with the depreciation formula.

$\begin{gathered} A=P(1-r)^t \\ \text{where;} \\ P\text{ is the original value} \\ r\text{ is the rate} \\ t\text{ is the time } \end{gathered}$

Hence,

$\begin{gathered} V=16300(0.94)^t \\ A=P(1-r)^t \\ \\ \text{Comparing both equations,} \\ P=16300 \\ 1-r=0.94 \\ 1-0.94=r \\ r=0.06 \\ To\text{ percentage,} \\ r=0.06\times100=6\text{ \%} \\ \\ \text{Hence, } \\ P\text{ is the purchase price} \\ r\text{ is the rate} \end{gathered}$

Therefore, the value of this car is decreasing at a rate of 6%. The purchase price of the car was $16,300.

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6 months ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .