Answer:
Explanation:
So you want to start by finding the area of the big square.
We know the width of the big square is 60cm. The length of the big square is
19
+
24
+
13
=
56
Big square
A
r
e
a
=
L
⋅
W
=
56
⋅
60
=
3360
Now let’s find the little square inside the bigger square. We see from the picture that the width is 34 and the length is 24cm. Using the same formula, we can find the area of the little square.
Little square
A
r
e
a
=
L
⋅
W
=
24
⋅
34
=
816
The area is enclosed in the area of the big square minus the area of the little square.
3360
−
816
=
2544
So the area enclosed is
2544
c
m
2
Step-by-step explanation:
Answer:
Below in bold.
Step-by-step explanation:
4x + y = 9
Let's put x = 0 then:
4(0) + y = 9
0 + y = 9
y = 9
So one. ordered pair that is a solution is (0, 9)
Compute the derivative dy/dx using the power, product, and chain rules. Given
x³ + y³ = 11xy
differentiate both sides with respect to x to get
3x² + 3y² dy/dx = 11y + 11x dy/dx
Solve for dy/dx :
(3y² - 11x) dy/dx = 11y - 3x²
dy/dx = (11y - 3x²)/(3y² - 11x)
The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when
11y - 3x² = 0
or
y = 3/11 x²
(provided that 3y² - 11x ≠ 0)
Substitute y into into the original equation:
x³ + (3/11 x²)³ = 11x (3/11 x²)
x³ + (3/11)³ x⁶ = 3x³
(3/11)³ x⁶ - 2x³ = 0
x³ ((3/11)³ x³ - 2) = 0
One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with
(3/11)³ x³ - 2 = 0
(3/11 x)³ - 2 = 0
(3/11 x)³ = 2
3/11 x = ³√2
x = (11•³√2)/3
Solving for y gives
y = 3/11 x²
y = 3/11 ((11•³√2)/3)²
y = (11•³√4)/3
So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).
