All categories

3 years ago

(1.2) to the third power (Explained)

Mathematics

1 answer:

katovenus [111]3 years ago

5 0

Answer: 1.2 to the third power is 1.728

Step-by-step explanation:

You might be interested in

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago

What is 119 over 8 in its simplest form?

ivolga24 [154]

119/8 is already in simplest form. However, if you attempt to change it to a mixed number the correct answer would be 14 7/8 (7 over 8).

Hope I helped!

3 0

3 years ago

The diameter of a circle measures 26 mm what is the circumference of the circle

nasty-shy [4]

The circumference is 81.68 mm. C= 2pi(r)

3 0

3 years ago

Read 2 more answers

Slope = 1, y-intercept = -4 ??

dalvyx [7]

Slope (m) = 1
Y-intercept (b) = -4

Formula: y = mx + b
To put it in the y=mx+b formula, you just have to plug it in so the answer would be...

y = 1x + -4

Hope that helped

8 0

3 years ago

Read 2 more answers

Nina and Jo both ran an 8 km race. Nina took 55 minutes to run the whole race. Jo started the race 3 minutes later than Nina but

grin007 [14]

Answer:

Step-by-step explanation:

First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is

$\frac{8km}{55min}=.145\frac{km}{min}$ and we can use that in a d = rt table:

d = r * t

Nina .145

Now we can fill in the distance which is 6 for both, since that is the distance where they met:

d = r * t

Nina 6 = .145

Jo 6 =

Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:

d = r * t

Nina 6 = .145 * t + 3

Jo 6 = r * t

As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then

6 = .145(t + 3) and

6 = .145t + .435 and

5.55 = .145t so

t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then

6 = r(38.379) and

r = .16 km/min

Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:

6 should equal .145454545( 38.37931034 + 3)

6 ?=? .145454545(41.37931034)

6 ?=? 6.0 so

Jo's rate is .16 km/min rounded

6 0

2 years ago