All categories

2 years ago

Where is Nathan173113?

Mathematics

1 answer:

Vinvika [58]2 years ago

3 0

Uhhhh not quite sure here

You might be interested in

1+1=<br> 2+2=<br> 3+3=<br> 4+4=<br> 5+5=

inysia [295]

1 + 1= 2

2 + 2 = 4

3 + 3 = 6

4 + 4 = 8

5 + 5 = 10

4 0

2 years ago

Read 2 more answers

What is a noun and adjective

7nadin3 [17]

Noun: A word that refers to something.

Adjective: A word that describes a trait.

8 0

1 year ago

Need help please I would really appreciate it

Nataly_w [17]

Answer:

The slope of line is 9

Step-by-step explanation:

We can use two pairs of input-output from the table to find the slope of the given line. Here x is input and y is output.

Slope is defined as the steepness of a line and is given by the formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Selected pairs from table are:

(-1,-18) and (0,-9)

Here

$x_1 = -1\\x_2 = 0\\y_1 = -18\\y_2 = -9$

Putting the values in the formula

$m = \frac{-9-(-18)}{0-(-1)}\\=\frac{-9+18}{0+1}\\= 9$

Hence,

The slope of line is 9

6 0

2 years ago

Evaluate the integral Integral from 0 to 1 Integral from 0 to 3 Integral from 3 y to 9 StartFraction 6 cosine (x squared )Over 5

Natalka [10]

Answer:

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$ $=\frac{18}{5}(1+\frac{sin2}{2})$

Step-by-step explanation:

cosine x²= cos x²

Rule

$\int x^ndx= \frac{x^{n+1}}{n+1}+c$

$\int cos \ mx \ dx = \frac{sin \ mx}{m}+c$
$\int \frac{1}{\sqrt x}dx = \frac{\sqrt x}{\frac{1}{2}} +c= 2\sqrt x+c$

Given that,

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$

$=\int ^1_0[\int^3_0(\int^9_{3y} \frac{6cos x^2}{5\sqrt z}dz)dy]dz$

$=\int^1_0[\int^3_0([\frac{6cos x^2 \times \sqrt z}{5\times \frac{1}{2}}]^9_{3y})dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( \sqrt 9-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( 3-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\frac{12cos x^2 \times( 3y-\frac{\sqrt{3}y^\frac{3}{2}}{\frac{3}{2}})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 3.3-\frac{2\sqrt{3}.3^\frac{3}{2}}{3})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 9-6)}{5}]dx$

$=\frac{18}{5}\int^1_02cos x^2dx$

$=\frac{18}{5}\int^1_0(1+cos2x)dx$

$=\frac{18}{5}[(x+\frac{sin2x}{2})]^1_0$

$=\frac{18}{5}(1+\frac{sin2}{2})$

6 0

2 years ago

Pls help will give crown to best answer!!

MissTica

Your answer is -B- i’m 100% sure

7 0

2 years ago