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.
This a pretty typical right triangle trig problem; the first step is to figure out what we have and what we want in relation to an acute angle in the problem.
Here we have a right triangle, G=90°, and we're given angle F=23°. So we have to name everything in relation to F.
31 = FG is <em>adjacent </em>to F.
x = GE is <em>opposite </em>to F.
OK, we have an opposite and adjacent; that tells us we need to use the tangent of F. Let's write it:
tan 23° = tan F = opp/adj = x/31
Solving,
x = 31 tan 23°
I hate the calculator part. I used to love that part.
x = 31 tan 23° ≈ 13.16 feet
Answer: 4) x ≈ 13.2 ft
Answer:
The right answer is 3,107.21
Step-by-step explanation:
the answer is thta
Vcylinder-Vsphere=water
Vcylinder=hpir^2
Vsphere=(4/3)pir^3
given
cylinderradius=30 and h=100
Vcylinder=100pi30^2=100pi900=90000pi
let's leave it in terms of pi for more exactness
sphereradius=12
Vsphere=(4/3)pi12^3=2304pi
cylinder-sphere=90000pi-2304pi=87696pi
using pi=3.14
water=275365.44 cubic cm
last option is corrrect
I will help you but I need to know what you need help with , find the area or volume?
