Answer:
(4,3,2)
Step-by-step explanation:
We can solve this via matrices, so the equations given can be written in matrix form as:
![\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%262%261%2620%5C%5C1%26-4%26-1%26-10%5C%5C2%261%262%2615%5Cend%7Barray%7D%5Cright%5D)
Now I will shift rows to make my pivot point (top left) a 1 and so:
![\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C2%261%262%2615%5C%5C3%262%261%2620%5Cend%7Barray%7D%5Cright%5D)
Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,
-2R1+R2=R2 , -3R1+R3=R3
![\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C0%269%264%2635%5C%5C0%2614%264%2650%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%2614%264%2650%5Cend%7Barray%7D%5Cright%5D)
4R2+R1=R1 , -14R2+R3=R3
![\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B7%7D%7B9%7D%26%5Cfrac%7B50%7D%7B9%7D%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%260%26-%5Cfrac%7B20%7D%7B9%7D%26-%5Cfrac%7B40%7D%7B9%7D%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B7%7D%7B9%7D%26%5Cfrac%7B50%7D%7B9%7D%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
, 
![\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%264%5C%5C0%261%260%263%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
Therefore the solution to the system of equations are (x,y,z) = (4,3,2)
Note: If answer choices are given, plug them in and see if you get what is "equal to". Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.
Answer:
146 miles
Step-by-step explanation:
hope this is correct
Answer:
-4
<em>BRAINLIEST, PLEASE!</em>
Step-by-step explanation:
-12/3 = -4
The angle formed by two tangles results in half of the difference of the two arcs.
That is to say
70 = .5(Major arc - minor arc)
Think of the major arc as the larger and the minor the smaller.
So therefore the difference in the two arcs = 140 (multiplying both sides of the above equation by 2.
140 = major arc - minor arc.
Another necessary idea is that the arc of a whole circle = 360 degrees, that is to say the major arc + minor arc = 360, giving us a second equation to solve.
Assuming x is the minor arc.
140 = major arc - x
360 = major arc + x
Use linear combination and add these equations to get
500 = 2*major arc
major arc = 250
minor arc or x = 110.
The answer is 110 degrees.
Answer:
1/32v²sin2θ
Step-by-step explanation:
Given the expression r(theta) = 1/16v²sinθcosθ
According to double angle of trigonometry identity;
Sin2θ = sin(θ+θ)
Sin2θ = sinθcosθ + cosθsinθ
Sin2θ = 2sinθcosθ
sinθcosθ = sin2θ/2 ... **
Substituting equation ** into the question
1/16v²sinθcosθ = 1/16v²(sin2θ/2)
1/16v²sinθcosθ = 1/2×1/16v²(sin2θ)
1/16v²sinθcosθ = 1/32v²sin2θ
Hence using the double angle identity, the equivalent expression is 1/32v²sin2θ
