Answer:
Ok
Step By Step:
thanks ?
Answer:
Add them up
Step-by-step explanation:
Answer:
The answer is below
Step-by-step explanation:
Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse 4x² + 16y² = 16
Solution:
Given that the ellipse has the equation: 4x² + 16y² = 16
let us make x the subject of the formula, hence:
4x² + 16y² = 16
4x² = 16 - 16y²
Dividing through by 4:
x² = (16 - 16y²)/4
x² = 4 - 4y²
Taking square root of both sides:
The points of the rectangle vertices is at (x,y), (-x,y), (x,-y), (-x,-y). Hence the rectangle has length and width of 2x and 2y.
The area of a rectangle inscribed inside an ellipse is given by:
Area (A) = 4xy
A = 4xy
Therefore the length = 2x = 2√2, the width = 2y = 2/√2
Answer:
7i sqrt(2)
Step-by-step explanation:
sqrt(-98)
We know that sqrt( ab) = sqrt(a) sqrt(b)
sqrt(-1) sqrt(98)
The sqrt (-1) = i
i sqrt(98)
i sqrt(49) sqrt(2)
i*7* sqrt(2)
7i sqrt(2)
Answer:
B: Julie has $150 to pay for cleaning services. She pays $8 for each hour of cleaning. She has $54 left afterwards.
Step-by-step explanation:
None