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3 years ago

Describe how to find the number of train tickets you can buy with $ 32.

Mathematics

2 answers:

Sloan [31]3 years ago

8 0

I think ur sop pose to pay the amount they tell u to pay . Hope it’s right good luck

Alex17521 [72]3 years ago

3 0

You would do $32/ the price of the train tickets.

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Find an exact value.

Westkost [7]

Answer:

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

Step-by-step explanation:

Convert the angle $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ to degrees:

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ .

Note, that $\left(-105^\circ\right)$ is the sum of two common angles: $\left(-45^\circ\right)$ and $\left(-60^\circ\right)$ .

$\displaystyle \cos\left(-45^\circ\right) = \cos\left(45^\circ\right) = \frac{\sqrt{2}}{2}$ .
$\displaystyle \cos\left(-60^\circ\right) = \cos\left(60^\circ\right) = \frac{1}{2}$ .

$\displaystyle \sin\left(-45^\circ\right) = -\sin\left(45^\circ\right) = -\frac{\sqrt{2}}{2}$ .
$\displaystyle \sin\left(-60^\circ\right) = -\sin\left(60^\circ\right) = -\frac{\sqrt{3}}{2}$ .

By the sum-angle identity of cosine:

$\cos(A + B) = \cos(A)\cdot \cos(B) - \sin(A) \cdot \sin(B)$ .

Apply the sum formula for cosine to find the exact value of $\cos\left(-105^\circ \right)$ .

$\begin{aligned}\cos\left(-105^\circ \right) &= \cos\left(\left(-45^\circ\right) + \left(-60^\circ\right)\right) \\ &= \cos\left(-45^\circ\right) \cdot \cos\left(-60^\circ\right)\right) - \sin\left(-45^\circ\right) \cdot \sin\left(-60^\circ\right)\right) \\ &= \frac{\sqrt{2}}{2} \times \frac{1}{2} - \left(-\frac{\sqrt{2}}{2}\right)\times \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\end{aligned}$ .

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ . In other words, $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ and $\left(-105^\circ\right)$ correspond to the same angle. Therefore, the cosine of $\displaystyle \left(-\frac{7\, \pi}{12}\right)\!$ would be equal to the cosine of $\left(-105^\circ\right)\!$ .

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \cos\left(-105^\circ\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

3 0

3 years ago

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. If (x + 2) is a fac

vitfil [10]

If x+2 is a factor of x^3+6x^2+kx+10, then if we replace x by -2 in the polynomial x^3-6x^2+kx+10, we must get zero:
x=-2→(-2)^3-6(-2)^2+k(-2)+10=0→
-8-6(4)-2k+10=0→
-8-24-2k+10=0
-2k-22=0
Solving for k. Adding 22 both sides of the equation:
-2k-22+22=0+22
-2k=22
Dividing both sides of the equation by -2:
-2k/(-2)=22/(-2)
k=-11

Answer: k = - 11

4 0

4 years ago

Please help me solve this

Aleks04 [339]

Answer:

Step-by-step explanation:

8 0

4 years ago

A unit rate is a rate in which A one of the quantities is 1. B the quantities add up to 1. C both of the quantities are 1. D nei

poizon [28]

Answer: A. because it is the only one that makes sense, the unit rate is where the second quantity is one unit: etc. 1 minute=60 seconds1 hour=60 minutes (or) 3600 seconds1 day=24 hours (or) 1440 minutes.

7 0

3 years ago

A cell phone company says that a fully charged battery lasts 14 hours. In reality, each battery behaves a little differently. Fo

gregori [183]

Answer:

The percentage is 50%

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 14 \ hours$

The standard deviation is $\sigma = 1 \ hour$

Generally on the normal distribution curve the mean divides the curve into two. with the half to the left of the mean (50% to the left )representing the percentage of the batteries that will last less than 14 hours and the area under the curve to the right (50% to the right)representing the percentage of batteries that will last longer than mean (14 hours )

Hence the percentage of the cell phone batteries that will be expected to last longer than 14 hours is 50%

8 0

4 years ago