A quadrilateral MUST be a parallelogram if it has one pair of opposite sides both parallel and congruent. A quadrilateral MUST be a parallelogram if it has both pairs of its opposite angles congruent (or equal in measure). A quadrilateral MUST be a parallelogram if it has both diagonals bisecting each other.

3 0

3 years ago

Y=x<br> y=-6x–14<br><br> (substitution method)

olga_2 [115]

Answer:

The solution is the point (-2,-2)

Step-by-step explanation:

we have

$y=x$ ----> equation A

$y=-6x-14$ ----> equation B

Solve the system by substitution

substitute equation A in equation B

$x=-6x-14$

Solve for x

$6x+x=-14\\7x=-14\\x=-2$

Find the value of y

substitute the value of x in the equation A

$y=x$

$y=-2$

therefore

The solution is the point (-2,-2)

3 0

4 years ago

Find the limit, if it exists. (if an answer does not exist, enter dne.) lim (x, y) → (0, 0) x4 − 20y2 x2 + 10y2

Firdavs [7]

Traveling along the x-axis, we have

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-20y^2}{x^2+10y^2}=\lim_{x\to0}\frac{x^4}{x^2}=\lim_{x\to0}x^2=0$

On the other hand, along the y-axis we get

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-20y^2}{x^2+10y^2}=\lim_{y\to0}\frac{-20y^2}{10y^2}=\lim_{y\to0}(-2)=-2$

Therefore the limit doesn't exist.

5 0

3 years ago

What is the slope of the line shown? need ASAP its due at 12 i will give brain list

Mamont248 [21]

Answer:

$\frac{1}{2} x$

Step-by-step explanation:

(1/2)x

use rise over run

how many units up/down and units right/left

Down and left means negative

5 0

3 years ago