$\left \{ {{x-2y=0} \atop {y=2x-3}} \right. \\\\Substitute\ second\ equation\ to\ first:\\\\x-2(2x-3)=0\\x-4x+6=0\\-3x=-6\ \ \ \ \ |:-3\\x=2\\\\y=2\cdot2-3\\y=4-3=1\\\\ \left \{ {{y=1} \atop {x=2}} \right.$

5 0

3 years ago

Read 2 more answers

Please please help. It’s due at 11:59. I don’t understand

Julli [10]

Answer:

Step-by-step explanation:

FYI,

C ∪ D = ( - ∞ , 2 ) ∪ [ 6 , ∞ ) 

C ∩ D = ∅ 

4 0

2 years ago

2) A student uses the substitution method to solve the system of equations below algebraically.

Rasek [7]

Pretty sure it's the last one.

5 0

3 years ago

Find the area of a circle with a radius of 3.

TiliK225 [7]

Answer:

A =28.26

Step-by-step explanation:

Using the area formula

A = pi r^2

We use pi = 3.14 and the radius is 3

A = 3.14 * (3)^2

A = 3.14 *9

A =28.26

5 0

3 years ago