Legs are the sides of an isosceles triangle that aren't the base, so in this case ZX would be the base. This rule is only for isosceles triangles. It doesn't apply to any other type of triangle.

Hope I helped, sorry if I'm wrong!

~Mschmindy

6 0

3 years ago

Read 2 more answers

Please help me.i don't get it.

alukav5142 [94]

Answer:

See below.

Step-by-step explanation:

The addition method is a good choice in this case since you have x in one equation and -x in the other equation, and x and -x add to zero, eliminating x.

x - 5.5y = -14

-x + 2.6y = 8.2

Add the equations. x and -x ad to zero, eliminating x. Then solve for y.

-2.9y = -5.8

Divide both sides by -2.9

y = 2

Substitute 2 for y in the first original equation, and solve for x.

x - 5.5y = -14

x - 5.5(2) = -14

x - 11 = -14

x = -3

Solution: x = -3; y = 2

Since the coefficients of y in the two original equations are not opposites, the addition method would not be the best method to use to solve for y first.

4 0

2 years ago

help ASAP thanks........

Fittoniya [83]

Answer: Yes, they are similar- and the Scale Factor is 2.5

Step-by-step explanation: To find the scale factor, simply divide the larger number on the bigger shape by its counterpart in the smaller image, so for example: 60 divided by 24 is 2.5, and there is your scale factor. You can further test this by multiplying 4 x 2.5 which gives us 10, the other value on the larger shape.

7 0

3 years ago

See the file below to see my question :)

RideAnS [48]

9514 1404 393

Answer:

a = 3
b = -18

Step-by-step explanation:

Using the least common denominator, we have ...

$2-\dfrac{x+1}{x-2}-\dfrac{x-4}{x+2}\\\\=\dfrac{2(x-2)(x+2)-(x+1)(x+2)-(x-4)(x-2)}{(x-2)(x+2)}\\\\=\dfrac{2(x^2-4)-(x^2+3x+2)-(x^2-6x+8)}{x^2-4}\\\\=\dfrac{(2-1-1)x^2+(-3+6)x+(-8-2-8)}{x^2-4}=\boxed{\dfrac{3x-18}{x^2-4}}$

The values of a and b are ...

a = 3

b = -18

3 0

3 years ago