I just did this question: Distance around a circle,C=nd,C=nr-in circumference of a circle
Area of a circle:A=nr,Measured in square units,(radius)(1/2 circumference)
Answer:
x=-12 (BUT IS EXTRENUS)
Step-by-step explanation:
If u dont know what extraneous is, dont worry about it...
Written as a power of ten is 256.×10^0 or 2.56×10^1
Answer:
y = 3
x = 2 - z
Step-by-step explanation:
We have the system:
2*x+y+2*z=7
2*x-y+2*z=1
5*x+y+5*z=13
In the first and second equations we have the term (2*x + 2*z) = A
Then we can rewrite the first two equations as:
A + y = 7
A - y = 1
isolating A in the first equation, we get:
A = 1 + y
Now we replace this in the other equation:
(1 + y) + y = 7
1 + 2*y = 7
2*y = 6
y = 3.
then:
A + y = 7
A + 3 = 7
A = 7- 3 = 4
A = 2*x + 2*z = 4.
Now let's go to the third equation:
(5*x + 5*z) + y = 13
we can rewrite the thing inside the parentheses as:
(5/2)*(2*x + 2*y) + y = 13
And we know that:
2*x + 2*z = 4
y = 3
then this can be written as:
(5/2)*(4) + 3 = 5*2 + 3 = 13
Then we can conclude that:
y = 3
2*z + 2*x = 4
2*(z + x) = 4
(z + x) = 4/2 = 2
x = 2 - z
Notice that the solution is not only a point, we have infinite solutions for this problem.
