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4 years ago

8

The proof shows that ABCD is a rectangle. Which of the following is the missing reason ?

1 answer:

Alik [6]4 years ago

6 0

The correct answer is: Opposite angles of a parallelogram are congruent.

In a parallelogram, the opposite angles are always congruent. The opposite angles are the angles diagonal across from each other.

You can tell by the order of ABCD that A and C are opposite. B and D are also opposite of each other and congruent.

You might be interested in

If s(x) = x – 7 and t(x) = 4x2 – x + 3, which expression is equivalent to (ToS) (x)

Gnoma [55]

In order to find the expression that is equivalent to (t*s)(x), use the following steps:
s(x) = x - 7t(x) = 4x^2 - x + 3
(t*s)(x) = t(s(x)) = t(x - 7) = 4(x - 7)^2 - (x - 7) + 3 = 4(x - 7)^2 - x + 7 + 3
The correct result would be 4(x – 7)2 – (x – 7) + 3.

5 0

3 years ago

Read 2 more answers

Just help me with 4 and 5 please‼️

Ray Of Light [21]

Answer:

Answer of 4 is 5 and answer of 5 is -1

4 0

4 years ago

The lengths of four insects are 0.02 inch,1/8 inch,and 2/3 inch list the lengths in inches from least to greatest

Alex Ar [27]

0.02
1/8 = 0.125
2/3 = 0.667

least to greatest : 0.02, 1/8, 2/3....although, u have only listed 3 insects, not 4

3 0

3 years ago

A: (0, –t)<br><br> B: (0, r)<br><br> C: (t, 0)<br><br> D: (–r, 0)

adell [148]

D should be the right answer

7 0

3 years ago

Suppose f is a function of one variable that has a continuous second derivative. Show that for any constants a and b, the functi

salantis [7]

Answer:

We can find the solution using chain rule,

For instance, if u(x,y) = f(ax+by), then

$u_{x}(x,y) = a f^{'} (ax+by)$ which represents derivative of x with respect to x

and then derivative of $u_{x}(x,y)$ with respect to y is

$u_{xy}(x,y) = (ab)^2 (f^{''} (ax+by))^2$ ,

Now, the derivative of $u_{x}(x,y)$ with respect to x, which is the second derivative, which is

$u_{xx}(x,y) = a^2 f^{''} (ax+by)$

and the derivative $u_{y}(x,y)$ and $u_{yy}(x,y)$ are

$u_{y}(x,y) = b f^{'} (ax+by)$ ,

$u_{yy}(x,y) = b^2 f^{''} (ax+by)$

Finally, the solution of PDE is

$u_{xx}(x,y) u_{yy}(x,y) - u_{xy} ^2 (x,y)$

$= (a^2 f^{''} (ax+by)) (b^2 f^{''} (ax+by)) - (ab)^2 (f^{''} (ax+by))^2$

$= (ab)^2 (f^{''} (ax+by))^2) - (ab)^2 (f^{''} (ax+by))^2$

= 0,

As the PDE is equal to 0, it means the function u(x,y) = f(ax+by) is the solution of the given PDE.

6 0

4 years ago