In △xyz, xy=xz. find the length of xy of △xyz if xy=2a, yz=3a+1, and xz=5a−12. 4 6.5 8 29 *q2: in △abc, m∠a=(4x+7)° and m∠b=(5x−3)°. what is m∠b? question 2 options: 47° 10° 45° 93°

XY=8

Step-by-step explanation:

In △XYZ, XY=XZ

It is isosceles triangle whose equal sides are XY=XZ

We are given,  

XY=2a

YZ=3a+1

XZ=5a-12

XY=XZ

So, 2a=5a-12

Solve for a,  

2a-5a=-12

-3a=-12

a=4

Now we put a=4 into XY=2a  

So, XY=2(4)=8

1.In △XYZ, XY=XZ.

We have to Find the length of XY of △XYZ if XY=2a, YZ=3a+1, and XZ=5a−12.

Now, X Y=X Z  [given]

→2 a= 5 a-12

Bringing like terms on same side

→2 a-5 a=-12

→-3 a=-12

→a=(-12)÷(-3)

a=4 is correct answer among all the options.

2.In △ABC, m∠A=(4x+7)° and m∠B=(5x−3)°.

What is m∠B?

Ans: as ∠ C not given we can't find ∠B.

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.