Fo a circle with radius, r and center (h,k)
te equation is
(x-h)²+(y-k)²=r²
d/2=r
8/2=4=r
center is (-2,-3)
(x-(-2))²+(y-(-3))²=4²
(x+2)²+(y+3)²=16
I use the substitution method, but you can also use elimination and matrix
1. Solve for y in 3x+y=14
y=14-3x
2. Substitute y=14-3x into x+2y=3
-5x+28=3
3. Solve for x in -5x+28=3
x=5
4. Substitute x=5 into y=14-3x
y=-1
5. Therefore,
x=5
y=-1
Have a nice day :D
Answer:
2 grams
Step-by-step explanation:
Answer:
1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x
Step-by-step explanation:
f(x) = sin (tan^-1 (ln(x)))
u substitution
d/du (sin u) * du /dx
cos (u) * du/dx
Let u =(tan^-1 (ln(x))) du/dx =d/dx (tan^-1 (ln(x)))
v substitution
Let v = ln x dv/dx = 1/x
d/dv (tan ^-1 v) dv/dx
1/( v^2+1) * dv/dx
=1/(ln^2x +1) * 1/x
Substituting this back in for du/dx
cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x
We know that cos (tan^-1 (a)) = 1/ sqrt(1+a^2)
cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x
1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x
Answer:
the fist one is A and the second one is D
Step-by-step explanation: