What is the inequality of 2/3 and 4/6?

Calculus The following curve passes through (3,1) Using linear lineralization

Novay_Z [31]

Answer:

Option 1.1

Step-by-step explanation:

The linearization of a curve implies the use of calculus to find the local value for the derivative and approximating the function by the use of the formula

$F(x) \approx F(x_0) + F'(x_0)(x-x_0)$

The function is given in such way that it's much easier to find the derivative by implicit differentiation than isolating any of the variables

$2x^2y+y=2x+13$

Differentiating with respect to x, we have

$4xy+2x^2y'+y'=2$

Computing y' in the given point (3,1) we have

4(3)(1)+2(9)y'+y'=2

$y'=\frac{2-12}{19}$

$y'=-\frac{10}{19}$

The function will be approximated with the expression

$F(x) = 1 -\frac{10}{19}(x-3)$

To find the approximate value for x=2.8

$F(2.8) = 1-\frac{10}{19}(-0.2)=1.1$

The correct value is the option 1.1

6 0

3 years ago

Why do people ...................

77julia77 [94]

Answer:

What?

Step-by-step explanation:

6 0

3 years ago

You lean a 30 foot ladder against your house and it reaches exactly to the top. The ladder makes a 24 degree angle with the grou

Furkat [3]

Sin 24 = h/30

h = 30 sin 24 = 12.2 ft

5 0

3 years ago

At the Olympic games, many events have several rounds of competition. One of these events is the men's 100100100-meter backstrok

padilas [110]

Answer:

C

Step-by-step explanation:

8 0

3 years ago

Please help. Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact valu

o-na [289]

Answer:

Area = (18 + 4.5π) cm²

Perimeter = (6√2 + 12 + 3π) cm

Step-by-step explanation:

The shaded region given is made up of triangle ABC and a semicircle

AB = BC = 6 cm (note that the triangle is a portion of a square)

Diameter of semi-circle (d) = BC = 6cm

Radius (r) = ½*6 = 3 cm

==>Area of the shaded region in terms of π

Area of shaded region = area of triangle + area of semicircle

Area = ½*a*b + ½*πr²

Area = ½*6*6 + ½*π3²

Area = 18 + ½*π9

Area = 18 + 4.5π

Area of the shaded portion = (18 + 4.5π) cm²

==>Perimeter of shaded region in terms of π

Perimeter of shaded region = perimeter of triangle + perimeter of semicircle

= Sum of all sides of the triangle + ½πd

Sides of triangles are AB = 6 cm, BC = 6 cm

Use Pythagorean theorem to find side AC:

AC² = AB² + BC²

AC² = 6² + 6² = 36 + 36 = 72

AC = √72 cm

Perimeter of shaded triangle = √72 + 6 + 6 = √72 + 12 = (6√2 + 12) cm

Perimeter of semicircle = ½*πd = ½π6

= 3π

Perimeter of the whole shaded region in terms of π = (6√2 + 12 + 3π) cm

3 0

3 years ago