All categories

3 years ago

How do you prove each of the following properties of a parallelogram?

Mathematics

1 answer:

Svet_ta [14]3 years ago

4 0

Step-by-step explanation:

the opposite angles are equal

i.e. i) < A = <C ii) < B = < D

the opposite sides are equal that is

i)AB =CD

ii) AD= BC

the diagonals bisect each other

i) AO=OC

ii)BO=OD

the adjacent angle are supplementary . that is

i)<D + < C=180 DEGREE

ii)<A+<D=180degree

iii)<B+<C=180 degree

You might be interested in

If r and s are positive integers, is \small \frac{r}{s} an integer? (1) Every factor of s is also a factor of r. (2) Every prime

Yuri [45]

Answer:

If statement(1) holds true, it is correct that $\small \frac{r}{s}$ is an integer.

If statement(2) holds true, it is not necessarily correct that $\small \frac{r}{s}$ is an integer.

Step-by-step explanation:

Given two positive integers $r$ and $s$ .

To check whether $\small \frac{r}{s}$ is an integer:

Condition (1):

Every factor of $s$ is also a factor of $r$ .

$r \geq s$

Let us consider an example:

$s = 5^2 \cdot 2\\r = 5^3 \cdot 2^2$

$\dfrac{r}{s} = \dfrac{5^3\cdot2^2}{5^2\cdot2} = 10$

which is an integer.

Actually, in this situation $s$ is a factor of $r$ .

Condition 2:

Every prime factor of s is also a prime factor of r.

(But the powers of prime factors need not be equal as we are not given the conditions related to powers of prime factors.)

Let

$r = 2^2\cdot 5\\s =2^4\cdot 5$

$\dfrac{r}{s} = \dfrac{2^3\cdot5}{2^4\cdot5} = \dfrac{1}{2}$

which is not an integer.

So, the answer is:

If statement(1) holds true, it is correct that $\small \frac{r}{s}$ is an integer.

If statement(2) holds true, it is not necessarily correct that $\small \frac{r}{s}$ is an integer.

8 0

3 years ago

I will give 30 points

user100 [1]

Answer:

1.7r

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

HELP ASAP

Naily [24]

Answer:

$y+4 = -\frac{1}{3} (x-1)$

Step-by-step explanation:

1) First, find the slope of the line. Use the slope formula $m = \frac{y_2-y_1}{x_2-x_1}$ to find the slope. Substitute the x and y values of the given points into the formula and solve:

$m = \frac{(-5)-(-4)}{(4)-(1)} \\m = \frac{-5+4}{4-1} \\m = \frac{-1}{3}$

So, the slope is $-\frac{1}{3}$ .

2) Use the point-slope formula $y-y_1 = m (x-x_1)$ to write the equation of the line in point-slope form. Substitute values for $m$ , $x_1$ , and $y_1$ into the formula.

Since $m$ represents the slope, substitute $-\frac{1}{3}$ in its place. Since $x_1$ and $y_1$ represent the x and y values of one point on the line, choose any one of the given points (it doesn't matter which one) and substitute the x and y values of it into the formula. (I chose (1,-4), as seen below). This gives the equation of the line in point-slope form.

$y-(-4) = -\frac{1}{3} (x-(1))\\y+4 = -\frac{1}{3} (x-1)$