Solving systems of equations X-2y=0 Y=2x-3

2 answers:

5 0

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

Musya8 [376]3 years ago

5 0

X - 2y = 0
y = 2x - 3

You subsitute. Since, y = 2x - 3, then you fill what y equals in for the y on the top equation.

x - 2 (2x - 3) = 0 :Distribute

x - 4x + 6 = 0 :Combine Like Terms

-3x + 6 = 0 :Subtract the 6 over to the 0
-6 -6

(-3x = -6) ÷ -3 :Now, divide the whole equation by -3

x = 2

Then, you want to get y, so you fill in the x that you got before into the bottom equation.

y = 2 ( 2) - 3 :Now, multiply 2 × 2 (Because of PEMDAS)

y = 4 - 3

y = 1

Your final answer is:

(2, 1)

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Suppose the radius of a circle is 3 units what is the circumference

HACTEHA [7]

Answer:

$6\pi$ or approximately $18.85$ units

Step-by-step explanation:

The formula for finding the circumference of a circle is $C=2\pi r$ , where $C$ = circumference and $r$ = radius. In this case, $r=3$ , so after plugging that into $C=2\pi r$ , we get:

$C=2\pi r\\ = 2*\pi *3\\=6\pi$

Since $\pi$ is approximately $3.1415$ , $6\pi =6*3.1415=18.849$ , which rounds to $18.85$ . Therefore, the circumference of the circle is $6\pi$ , or approximately $18.85$ units. Hope this helps!