Answer:
The number of machines to complete job in 2 hours is 6.
Step-by-step explanation:
Using the relation of the rate of each one machine and knowing that adding up both rates.
Rate R = 
'x' is the job
Rate S = 
'x' is the job
So, using the equation of rates
Rate total is going to be a job in 2 hours so = 
<em>adding up</em>
<em>factoring the x</em>
<em>simplifying</em>
x= 
x= 6
2/X^2=4/X+1 times by both denominators
2(x+1)=4(x^2)
2x+2=4x^2
4x^2-2x-2=0
(2x+1)(2x-2)=0
x=2x+1 or 2x-2
x=-1/2 or 1
From what I'm understanding of these questions, the biggest thing you need to answer these is the formulas for cylinders and triangular prisms. I'm not sure what the quantities are for either question so I'm going to work with made up numbers to give examples for the formulas. For number 2 with the cylinder, let's consider the formula first:
π × r2 × h <em>OR </em>pi (3.14) times radius squared times height
If you have the height and you have pi, all you need to take is the doubled radius (aka multiply it by 2) and plug that back into the formula. For the sake of an example, I'm going to make up the number 2 for the radius and 6 for the height. Here's what that would look like:
r = 2; double it, resulting in 4
pi x 4^2 x 6
3.14 x 16 x 6
= 301.44
Work with the actual numbers you have and you're good to go.
For number 3, reducing something by 1/2 means dividing by 2. Let's consider the formula and then work through another example:
1/2 x b x h x l <em>OR </em> 1/2 times base times height times length
For the sake of an example, I'll use 10 for the height, 15 for the base, and 20 for the length:
h = 10; reduce by 1/2, resulting in 5
1/2 x 15 x 5 x 20
= 750
Plug in your actual quantities, and remember your volume units. Hope this helps!
We know that:
There is also an interesting property that relates the sine and the cosine of an angle:
We can find the cosine of theta using this equation:
![\begin{gathered} \cos ^2(\theta_1)=1-\sin ^2(\theta_1) \\ \cos (\theta_1)=\sqrt{1-\sin^2(\theta_1)} \\ \cos (\theta_1)=\sqrt[]{1-(-\frac{12}{13})^2} \\ \lvert\cos (\theta_1)\rvert=\sqrt[]{1-\frac{144}{169}}=\sqrt[]{\frac{25}{169}} \\ \lvert\cos (\theta_1)\rvert=\frac{5}{13} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%5E2%28%5Ctheta_1%29%3D1-%5Csin%20%5E2%28%5Ctheta_1%29%20%5C%5C%20%5Ccos%20%28%5Ctheta_1%29%3D%5Csqrt%7B1-%5Csin%5E2%28%5Ctheta_1%29%7D%20%5C%5C%20%5Ccos%20%28%5Ctheta_1%29%3D%5Csqrt%5B%5D%7B1-%28-%5Cfrac%7B12%7D%7B13%7D%29%5E2%7D%20%5C%5C%20%5Clvert%5Ccos%20%28%5Ctheta_1%29%5Crvert%3D%5Csqrt%5B%5D%7B1-%5Cfrac%7B144%7D%7B169%7D%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B25%7D%7B169%7D%7D%20%5C%5C%20%5Clvert%5Ccos%20%28%5Ctheta_1%29%5Crvert%3D%5Cfrac%7B5%7D%7B13%7D%20%5Cend%7Bgathered%7D)
Since theta is in the third quadrant then its cosine must be a negative number so:
