Will the opposite of a number always sometimes or never be greater than the number itself

1 answer:

7 0

Sometimes, because if the number is negative, the opposite is greater, but if the number is positive, the opposite is smaller.

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Multiple-Choice Integration, Picture Included, Please Include Work

ValentinkaMS [17]

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx$

Recall that a circle of radius 2 centered at the origin has equation

$x^2+y^2=4\implies y=\pm\sqrt{4-x^2}$

where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius $r$ is $\pi r^2$ , it follows that the area of a quarter-circle would be $\dfrac{\pi r^2}4$ .

You have $r=2$ , so the definite integral is equal to $\dfrac{2^2\pi}4=\pi$ .

Another way to verify this is to actually compute the integral. Let $x=2\sin u$ , so that $\mathrm dx=2\cos u\,\mathrm du$ . Now

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du$

Recall the half-angle identity for cosine:

$\cos^2u=\dfrac{1+\cos2u}2$

This means the integral is equivalent to

$\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi$

4 0

3 years ago

if a piece of string which is 368cm long must be cut into 8 equal pieces how long will each piece be (answer in mm)

Maksim231197 [3]

1cm= 10mm
So you do 368 divided by 80 which is 4.6mm

3 0

3 years ago

Consider the matrix shown below:

velikii [3]

Answer:

$A)\ \ \ \ \left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

Step-by-step explanation:

Given the matrix: $\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right]$ , it's inverse is calculated using the formula:

$\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]^{-1}=\frac{1}{det\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] }\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]$

#Therefore, we calculate as;

$\frac{1}{det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right] }\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right] \\\\\\\\\#det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right] =1\\\\\\\\=\frac{1}{1}\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right] \\\\\\\\=\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$