Answer:
Equation of the circle =
80 = ( x - 1)^2 + (y + 1)^2
Step-by-step explanation:
Center = (1 , -1)
Point = ( 5 , 7)
Eqn of a circle is
r^2 = (x - h)^2 + (y - k)^2
We are not given the radius of the circle, fortunately we are provided the information that the circle contains a point ( 5, 7), so we use the above information to find r
Using the center (1, -1)
h - 1
k - -1
r^2 = (x - 1)^2 + ( y - -1)^2
r^2 = (x - 1)^2 + (y + 1)^2
With the point ( 5,7)
x = 5
y = 7
r^2 = ( 5 - 1)^2 + ( 7 + 1)^2
= 4^2 + 8^2
= 16 + 64
= 80
r^2 = 80
r = square root of 80
r = 8.94
The r = 8.94 , which means the equation of the circle is
80 = (x - 1)^2 + ( y + 1)^2
Answer:
2.45c + 1.65c = 4.12 + 0.75
Step-by-step explanation:
To write an equation to find the value for c, we need to declare what c is first.
c = price of fruit
2.45c + 1.65c = 4.12 + 0.75
Now we multiplied c to 2.45 and 1.65 and added them together, because whatever the value of c is will give us the equivalence of the sum of 4.12 + 0.75.
Now to check if the equation is right, let's solve for c.
2.45c + 1.65c = 4.12 + 0.75
4.1c = 4.87
Now to get the value of c, we divide both sides of the equation by 4.1.

c = 1.19
Now let's substitute the value of c in the equation to see if we got it right.
2.45(1.19) + 1.65(1.19) = 4.12 + 0.75
2.92 + 1.96 = 4.87
4.87 = 4.87
Therefore concluding that the value of c is 1.19.
Lagrange multipliers:







(if

)

(if

)

(if

)
In the first octant, we assume

, so we can ignore the caveats above. Now,

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of

.
We also need to check the boundary of the region, i.e. the intersection of

with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force

, so the point we found is the only extremum.
3x - 4y = 9 and 2x + 3y = 7, which of the following would be the best method?
Answer:
k ≥2
Step-by-step explanation:
k+ 1/3≥7/3
Subtract 1/3 from each side
k+ 1/3-1/3≥7/3-1/3
k ≥6/3
k ≥2