The height of mast is 19.21 feet
<em><u>Solution:</u></em>
Given that, 25-ft rope is stretched from the top of a ship's mast to a cleat on the deck
The cleat is 16 ft from the base of the mast
To find: height of mast
The ship mast, deck and base forms a right angled triangle
The figure is attached below
AC = length of rope = 25 feet
BC = distance between base of mast and cleat = 16 feet
AB = height of mast = ?
Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle
By above definition for right angled triangle ABC,
![AC^2=AB^2+BC^2](https://tex.z-dn.net/?f=AC%5E2%3DAB%5E2%2BBC%5E2)
![25^2=AB^2+16^2](https://tex.z-dn.net/?f=25%5E2%3DAB%5E2%2B16%5E2)
![625 = AB^2+256\\\\AB^2=625-256\\\\AB^2=369\\\\\text{Take square root on both sides }\\\\AB = \sqrt{369}\\\\AB = 19.21](https://tex.z-dn.net/?f=625%20%3D%20AB%5E2%2B256%5C%5C%5C%5CAB%5E2%3D625-256%5C%5C%5C%5CAB%5E2%3D369%5C%5C%5C%5C%5Ctext%7BTake%20square%20root%20on%20both%20sides%20%7D%5C%5C%5C%5CAB%20%3D%20%5Csqrt%7B369%7D%5C%5C%5C%5CAB%20%3D%2019.21)
Thus height of mast is 19.21 feet
Answer:
8192
Step-by-step explanation:
The answer is <em>8192</em> but in your multiple-choice problem it's ![2^{13}](https://tex.z-dn.net/?f=2%5E%7B13%7D)
Answer:
The travel size costs 6 % as much as the regular size. Thus the regular size would be more economical and thus a better buy
Step-by-step explanation:
step 1
<em>Find out the unit rate of a regular tube of toothpaste</em>
we know that
To find out the unit rate, divide the total cost by the total weight
so
![\frac{2.50}{3.2}=\$0.78\ per\ ounce](https://tex.z-dn.net/?f=%5Cfrac%7B2.50%7D%7B3.2%7D%3D%5C%240.78%5C%20per%5C%20ounce)
step 2
<em>Find out the unit rate of a travel size tube of toothpaste</em>
we know that
To find out the unit rate, divide the total cost by the total weight
so
![\frac{1.00}{1.2}=\$0.83\ per\ ounce](https://tex.z-dn.net/?f=%5Cfrac%7B1.00%7D%7B1.2%7D%3D%5C%240.83%5C%20per%5C%20ounce)
step 3
Find out the percent of increase
we have that
The unit rate
represent the 100%
so by proportion
Find out what percentage represent the difference ($0.83-$0.78=$0.05 per ounce)
![\frac{0.83}{100\%}=\frac{0.05}{x}\\\\x=100(0.05)/0.83\\\\x=6\%](https://tex.z-dn.net/?f=%5Cfrac%7B0.83%7D%7B100%5C%25%7D%3D%5Cfrac%7B0.05%7D%7Bx%7D%5C%5C%5C%5Cx%3D100%280.05%29%2F0.83%5C%5C%5C%5Cx%3D6%5C%25)
therefore
The travel size costs 6 % as much as the regular size. Thus the regular size would be more economical and thus a better buy
Answer:
<h3> 1.)
![x_1=\dfrac{7+\sqrt{13}}{2}\,,\qquad x_2=\dfrac{7-\sqrt{13}}{2}](https://tex.z-dn.net/?f=x_1%3D%5Cdfrac%7B7%2B%5Csqrt%7B13%7D%7D%7B2%7D%5C%2C%2C%5Cqquad%20x_2%3D%5Cdfrac%7B7-%5Csqrt%7B13%7D%7D%7B2%7D)
</h3><h3> 2.)
![x_1=-\dfrac13\,,\qquad x_2=-3](https://tex.z-dn.net/?f=x_1%3D-%5Cdfrac13%5C%2C%2C%5Cqquad%20x_2%3D-3)
</h3>
Step-by-step explanation:
1.)
![x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b=-7\,,\ c = 9\\\\\\x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{7\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}{2}\\\\x_1=\dfrac{7+\sqrt{13}}{2}\,,\qquad x_2=\dfrac{7-\sqrt{13}}{2}](https://tex.z-dn.net/?f=x%5E2%20-%207x%20%2B%209%20%3D%200%5Cquad%5Cimplies%5Cquad%20a%3D1%5C%2C%2C%5C%20b%3D-7%5C%2C%2C%5C%20c%20%3D%209%5C%5C%5C%5C%5C%5Cx%3D%5Cdfrac%7B-b-%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B7%5Cpm%5Csqrt%7B%28-7%29%5E2-4%5Ccdot1%5Ccdot9%7D%7D%7B2%5Ccdot1%7D%3D%5Cdfrac%7B7%5Cpm%5Csqrt%7B49-36%7D%7D%7B2%7D%5C%5C%5C%5Cx_1%3D%5Cdfrac%7B7%2B%5Csqrt%7B13%7D%7D%7B2%7D%5C%2C%2C%5Cqquad%20x_2%3D%5Cdfrac%7B7-%5Csqrt%7B13%7D%7D%7B2%7D)
2.)
![3x^2 + 10x = -3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b=10\,,\ c = 3\\\\\\x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}=\dfrac{-10\pm\sqrt{100-36}}{6}=\dfrac{-10\pm\sqrt{64}}{6}\\\\x_1=\dfrac{-10+8}{6}=\dfrac{-2}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}{6}=\dfrac{-18}{6}=-3](https://tex.z-dn.net/?f=3x%5E2%20%2B%2010x%20%3D%20-3%5C%5C%5C%5C3x%5E2%2B10x%2B3%3D0%5Cquad%5Cimplies%5Cquad%20a%3D3%5C%2C%2C%5C%20b%3D10%5C%2C%2C%5C%20c%20%3D%203%5C%5C%5C%5C%5C%5Cx%3D%5Cdfrac%7B-b-%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B-10%5Cpm%5Csqrt%7B10%5E2-4%5Ccdot3%5Ccdot3%7D%7D%7B2%5Ccdot3%7D%3D%5Cdfrac%7B-10%5Cpm%5Csqrt%7B100-36%7D%7D%7B6%7D%3D%5Cdfrac%7B-10%5Cpm%5Csqrt%7B64%7D%7D%7B6%7D%5C%5C%5C%5Cx_1%3D%5Cdfrac%7B-10%2B8%7D%7B6%7D%3D%5Cdfrac%7B-2%7D6%3D-%5Cdfrac13%5C%2C%2C%5Cqquad%20x_2%3D%5Cdfrac%7B-10-8%7D%7B6%7D%3D%5Cdfrac%7B-18%7D%7B6%7D%3D-3)
Answer:
![(x,y) = (r,\theta)\\](https://tex.z-dn.net/?f=%28x%2Cy%29%20%3D%20%28r%2C%5Ctheta%29%5C%5C)
i)![(4,-4) = (\pm 4\sqrt{2},\dfrac{7 \pi}{4})\\](https://tex.z-dn.net/?f=%284%2C-4%29%20%3D%20%28%5Cpm%204%5Csqrt%7B2%7D%2C%5Cdfrac%7B7%20%5Cpi%7D%7B4%7D%29%5C%5C)
ii) ![(-1, \sqrt{3}) = (\pm 2,\dfrac{2 \pi}{4})\\](https://tex.z-dn.net/?f=%28-1%2C%20%5Csqrt%7B3%7D%29%20%3D%20%28%5Cpm%202%2C%5Cdfrac%7B2%20%5Cpi%7D%7B4%7D%29%5C%5C)
Thinking Process:
This can either be solved using complex variables or simply trigonometry (they are all the same in a way)
![(a,b) = a + ib = r(\cos{(\theta)} + i\sin{(\theta)}) = r e^{(i \theta)} = r\angle{\theta}](https://tex.z-dn.net/?f=%28a%2Cb%29%20%3D%20a%20%2B%20ib%20%3D%20r%28%5Ccos%7B%28%5Ctheta%29%7D%20%2B%20i%5Csin%7B%28%5Ctheta%29%7D%29%20%3D%20r%20e%5E%7B%28i%20%5Ctheta%29%7D%20%3D%20r%5Cangle%7B%5Ctheta%7D)
But we'll simply go with trigonometry and good old Pythagorean theorem.
Solution:
For Cartesian coordinates are: ![(x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29)
To covert them into ![r \angle{\theta}](https://tex.z-dn.net/?f=r%20%5Cangle%7B%5Ctheta%7D)
We can use:
- The Pythagoras theorem to find
![r](https://tex.z-dn.net/?f=r)
![r = \pm \sqrt{x^2 + y^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D)
- Arctangent to find
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%7D)
These are related since:
![x = r\cos{(\theta)}](https://tex.z-dn.net/?f=%20x%20%3D%20r%5Ccos%7B%28%5Ctheta%29%7D)
![y = r\sin{(\theta)}](https://tex.z-dn.net/?f=%20y%20%3D%20r%5Csin%7B%28%5Ctheta%29%7D)
So, let's start solving:
Part a) (4,-4)
For r:
![r = \pm \sqrt{x^2 + y^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D)
![r = \pm \sqrt{4^2 + (-4)^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7B4%5E2%20%2B%20%28-4%29%5E2%7D)
![r = \pm \sqrt{32}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7B32%7D)
![r = \pm 4 \sqrt{2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%204%20%5Csqrt%7B2%7D)
i) for
we'll go with
ii) for
we'll go with
For theta:
![\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%7D)
![\theta = \tan^{-1}{\left(\dfrac{-4}{4}\right)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28%5Cdfrac%7B-4%7D%7B4%7D%5Cright%29%7D)
![\theta = \tan^{-1}{(-1)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%7B%28-1%29%7D)
![\theta = -45\texdegree \,\, or -\dfrac{\pi}{4} radians](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20-45%5Ctexdegree%20%5C%2C%5C%2C%20or%20-%5Cdfrac%7B%5Cpi%7D%7B4%7D%20radians)
One thing to keep in mind is that the point (4,-4) lies in the 4th quadrant of the xy-plane, and the range given for
is
. So we have to add
to our answer (we have gone around the circle and stopped at the same place as
radians)
radians
radians
The Answer for part(a) is ![(\pm 4\sqrt{2}, \dfrac{7 \pi}{4})](https://tex.z-dn.net/?f=%28%5Cpm%204%5Csqrt%7B2%7D%2C%20%5Cdfrac%7B7%20%5Cpi%7D%7B4%7D%29)
Part b) (-1, sqrt(3))
For r:
![r = \pm \sqrt{x^2 + y^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D)
![r = \pm \sqrt{(-1)^2 + (\sqrt{3})^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7B%28-1%29%5E2%20%2B%20%28%5Csqrt%7B3%7D%29%5E2%7D)
![r = \pm \sqrt{(-1)^2 + (\sqrt{3})^2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%20%5Csqrt%7B%28-1%29%5E2%20%2B%20%28%5Csqrt%7B3%7D%29%5E2%7D)
i)
for ![r > 0](https://tex.z-dn.net/?f=r%20%3E%200%20)
ii)
for
For theta:
![\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%7D)
![\theta = \tan^{-1}{\left(\dfrac{\sqrt{3}}{-1}\right)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B-1%7D%5Cright%29%7D)
this not in range of theta:
.
and the point
lies in the 2nd quadrant, so we can add
radians to our answer
in range: ![0 \leq \theta \leq 2 \pi](https://tex.z-dn.net/?f=0%20%5Cleq%20%5Ctheta%20%5Cleq%202%20%5Cpi)
![\theta = \dfrac{2 \pi}{3}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cdfrac%7B2%20%5Cpi%7D%7B3%7D%20)
The Answer for part(b) is ![(\pm 2, \dfrac{2 \pi}{3})](https://tex.z-dn.net/?f=%28%5Cpm%202%2C%20%5Cdfrac%7B2%20%5Cpi%7D%7B3%7D%29)