All categories

3 years ago

Oliver read for 450450450 minutes this month. His goal is to read for 10\%10%10, percent more minutes next month.

Mathematics

1 answer:

OLEGan [10]3 years ago

7 0

Answer: Oliver's goal is to read 45 minutes more next month

Step-by-step explanation:

Firstly, we need to know how much in minutes is the $10\%$ of $450 min$ :

$450 min(10\%)=(450 min)(\frac{10}{100})=45 min$

This means Oliver wants to read 45 minutes more next month. If we add this number to the 450 minutes he read this month, we have a total of 495 minutes.

You might be interested in

HELP PLEASE ! TRUE OR FALSE ?!

miss Akunina [59]

This is false because on 2) x does not equal 3.

8 0

4 years ago

Sin(x)^4+ cos(x)^4=1/2

posledela

I'll assume that what was meant was $\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$ .

The exponent in the funny place is just an abbreviation: $\sin ^4 x = (\sin x)^4$ .

I hope that's what you meant. Let me know if I'm wrong.

Let's start from the old saw

$\cos^2 x + \sin ^2x = 1$

Squaring both sides,

$(\cos^2 x + \sin ^2x)^2 = 1^2$

$\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1$

$\cos^4 x + \sin ^4x = 1 - 2 \cos ^2 x \sin ^2x$

So now the original question

$\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$

becomes
$1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}$

$4 \cos ^2 x \sin ^2x = 1$

Now we use the sine double angle formula

$\sin 2x = 2 \sin x \cos x$

We square it to see

$\sin^2 2x = 4\sin^2 x \cos^2 x = 1$

Taking the square root,

$\sin 2x = \pm 1$

Not sure how you want it; we'll do it in degrees.

When we know the sine of an angle, there's usually two angles on the unit circle that have that sine. They're supplementary angles which add to $180^\circ$ . But when the sine is 1 or -1 like it is here, we're looking at $90^\circ$ and $-90^\circ$ , which are essentially their own supplements, slightly less messy.

That means we have two equations:

$\sin 2x = 1 = \sin 90^\circ$

$2x = 90^\circ + 360^\circ k \quad$ integer $k$

$x = 45^\circ + 180^\circ k$

or

$\sin 2x = -1 = \sin -90^\circ$

$2x = -90^\circ+ 360^\circ k$

$x = - 45^\circ + 180^\circ k$

We can combine those for a final answer,

$x = \pm 45^\circ + 180^\circ k \quad$ integer $k$

Check. Let's just check one, how about

$x=-45^\circ + 180^\circ = 135^\circ$

$\sin(135)= 1/\sqrt{2}$

$\sin ^4(135)=(1/\sqrt{2})^4 = 1/4$

$\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4$

$\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark$

6 0

3 years ago

Help someoneee!!!! <br> ill give a brainliest

Serjik [45]

Answer:

-13.5

Step-by-step explanation:

-4/3w=18

You would need to get the w by itself on one side so the easiest way to do so would be to divide both sides my -4/3.

This would give you an answer of w= -27/2 or w= -13.5

8 0

3 years ago

Calculus 3 chapter 16

o-na [289]

Evaluate $\vec F$ at $\vec r$ :

$\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k$

Compute the line element $d\vec r$ :

$d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt$

Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$

3 0

2 years ago

0.04(y - 9) +0.16y = 0.12y-0.5<br> y=<br> (Type an integer or a decimal.)

Vladimir [108]

The answer is going to be y=-1.75 hope this helps!!

3 0

3 years ago