Your answer would be option B. 2y² - y - 6 = 0. This is because if you were to substitute x = y² - 1 into the equation 2x - y = 4, you would get 2(y² - 1) - y = 4, which expands into 2y² - 2 - y = 4, and then simplifies to 2y² - y - 6 = 0.
I hope this helps!
m x H = ![\left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%2637.5%26-12.5%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Step 1; Multiply 5 with this matrix
and we get a matrix ![\left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%2610%5C%5C20%2640%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Multiply the fraction
with the matrix
and we get ![\left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2m%7D%7B5%7D%20%26%5Cfrac%7B4m%7D%7B5%7D%20%5C%5C%5Cfrac%7B8m%7D%7B5%7D%20%26%5Cfrac%7B16m%7D%7B5%7D%20%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step2; Now equate corresponding values of the matrices with each other.
-5 =
and so on. By equating we get the value of m as 
Step 3; Add the matrices to get the value of matrix m.
Adding the three matrices on the RHS we get
.
Step 4; Adding the matrices on the LHS we get the resulting matrix as H +
. Equating the matrices from step 3 and 4 we get the value of H as ![\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%26-1%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step 5; Now to find the value of m x H we need to multiply the value of
with the matrix
Step 6; Multiplying we get the matrix m x H = [ -25
]
Answer:
66 sweets
Step-by-step explanation:
all you need to do is multiply 11 by 6 to get your answer seeing that ⅙ of the jar is 11
Same slope, different y-intercept
Answer:
327.4
Step-by-step explanation:
All you have to do to simplify a number written in scientific notation is to move the decimal place to the left (if the number is negative) and to the right (if the number is positive).