The equation of variation in which y varies jointly as x and z and inversely as the product of w and p is y=0.5(xz/wp).
Given that variable y varies jointly as x and z and inversely as the product of w and p,where y=7/28 where x=7,z=4,w=7 and p=8.
We are required to find the equation of variation.
To solve this problem we must apply the following procedure:
1) We have that y varies jointly as x and z and inversely as the product of w and p. Therefore we can write the following equation,where k is the constant of proportionality:
y=k(xz/wp)----------1
Now we have to solve for the constant of proportionality as done under:
k=ywp/xz-------------2
Using the values in equation 1.
k=(7/28)(7)(8)/(7)(4)
=0.5
Using all the values in the equation 2.
y=0.5(xz/wp)
Hence the equation of variance is y=0.5(xz/wp).
Question is incomplete.The following values should be included:
y=7/28 where x=7,z=4,w=7 and p=8.
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(2,1) i believe (?)
(4,5) (0,-3)
you add the x’s so the 4 and the 0 which is 4 then you divide it by 2 since there are two x numbers. for the y you would do the same 5+-3 ( plus and a minus makes a minus, 5-3) which is 2 then divide it by 2 again. Finally you get (2,1)
Answer:
4x-2y
Step-by-step explanation:
Sides:
a = 19 m
b = 25 m
c = 31 m
Angles:
A = 37.7843 °
B = 53.7236 °
C = 88.492 °
Other:
P = 75 m
s = 37.5 m
K = 237.418 m2
r = 6.33114 m
R = 15.5054 m
Answer:
x=4, y=4, λ=-16
Step-by-step explanation:
We have this 3x3 system of linear equations:
λ
λ
So, let's rewrite the system in its augmented matrix form
![\left[\begin{array}{cccc}4&0&1&0\\0&4&1&0\\1&1&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D4%260%261%260%5C%5C0%264%261%260%5C%5C1%261%260%268%5Cend%7Barray%7D%5Cright%5D)
Let´s apply row reduction process to its associated augmented matrix:
Swap R1 and R3
![\left[\begin{array}{cccc}1&1&0&8\\0&4&1&0\\4&0&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%260%268%5C%5C0%264%261%260%5C%5C4%260%261%260%5Cend%7Barray%7D%5Cright%5D)
R3-4R1
![\left[\begin{array}{cccc}1&1&0&8\\0&4&1&0\\0&-4&1&-32\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%260%268%5C%5C0%264%261%260%5C%5C0%26-4%261%26-32%5Cend%7Barray%7D%5Cright%5D)
R3+R2
![\left[\begin{array}{cccc}1&1&0&8\\0&4&1&0\\0&0&2&-32\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%260%268%5C%5C0%264%261%260%5C%5C0%260%262%26-32%5Cend%7Barray%7D%5Cright%5D)
Now we have a simplified system:
x+y+0=0
0+4y+λ=0
0+0+2λ=-32
Solving for λ, x, and y
λ=-16
x=4
y=4
