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3 years ago

Which counterexample shows that the conjecture "Every parallelogram is also a rectangle" is false?

Mathematics

2 answers:

mafiozo [28]3 years ago

5 0

A parallelogram without four right angles would prove that to be false.

Ivanshal [37]3 years ago

3 0

Yep that'd be false since they are also rectangles

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HELP NEEDED 20 POINTS AND BRAINLIST!!!! Please explain your answer and no links! please!

guajiro [1.7K]

Answer:

From (0,0) and go down 1 and put a dot then down 4 and over to the right 1 and draw a line.

Step-by-step explanation:

Hope this helps have a nice day!

6 0

2 years ago

Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is

ohaa [14]

The vectors in $S$ form a basis of $P_2$ if they are mutually linearly independent and span $P_2$ .

To check for independence, we can compute the Wronskian determinant:

$\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0$

The determinant is non-zero, so the vectors are indeed independent.

To check if they span $P_2$ , you need to show that any vector in $P_2$ can be expressed as a linear combination of the vectors in $S$ . We can write an arbitrary vector in $P_2$ as

$p=ax^2+bx+c$

Then we need to show that there is always some choice of scalars $k_1,k_2,k_3$ such that

$k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p$

This is equivalent to solving

$(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c$

or the system (in matrix form)

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

This has a solution if the coefficient matrix on the left is invertible. It is, because

$\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0$

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}$

Then

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}$

$\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}$

A solution exists for any choice of $a,b,c$ , so the vectors in $S$ indeed span $P_2$ .

The vectors in $S$ are independent and span $P_2$ , so $S$ forms a basis of $P_2$ .

5 0

3 years ago

Hi, the answer is K but can anyone show why

dlinn [17]

Let's do 51 and 52.

51. The contrapositive has the same truth value as the original statement. That's opposed to the converse, which may or may not be true independent of the original statement.

The contrapositive of IF P THEN Q is IF not Q THEN not P. They're equivalent. Here that's If the cat is not female then it is not tricolor.

Answer: C

52.

$(x^3)^{(4-b^2)}=1$

$x^{3(4-b^2)} = 1$

For the statement to be true, the exponent must be zero:

$3(4-b^2) = 0$

$b^2 = 4$

$b = \pm 2$

Both positive 2 and negative 2 have a square of 4.

Answer: K

By the way, usually we assume $0^0=1$ so the restriction that $x \ne 0$ isn't really necessary. Think of the definition of a polynomial or the binomial expansion: