HETY is a parallelogram.
HT and EY are diagonals. We know that diagonals divides the parallelogram into two equal parts.
So ar(HET) = ar(HTY)
And, ar(HEY) = ar(EYT) now, in AHET, diagonal EY bisects the line segment HT and also the AHET,
∴ar(AHOE) = ar(AEOT)
Similarly in AETY
ar(ΔΕΟΤ) = ar(ΔΤΟΥ)
And in AHTY,
ar(ATOY) = ar(AHOY)
That means diagonals in parallelogram divides it into four equal parts.
Hence Proofed.
![\sqrt[8]{48^4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B8%5D%7B48%5E4%7D%20)
can be rewritten as
to the
power:
Now, applying exponent rules (multiply exponent inside parentheses by the one outside parentheses), we get:
This is equivalent to
, so now, we just simplify:
So:
![\sqrt[8]{48^4} =4 \sqrt{3}](https://tex.z-dn.net/?f=%20%5Csqrt%5B8%5D%7B48%5E4%7D%20%3D4%20%5Csqrt%7B3%7D)
The answer is D.
You have to make the whole numbers equal to then solve for the exponents:
F(x) means y, therefore, y is 9.
9 = 3^x.
3^2 is the same as 9.
Then, 3^2 = 3^x.
Then solve for the exponents,
2 = x
Therefore your answer is 2.
Answer:
(2.83 , 1 , 4)
Step-by-step explanation:
Rewrite these equations in matrix form
![\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%26-1%5C%5C4%26-2%26-2%5C%5C3%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
we can write it like this,
so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.
We get the inverse of matrix A,
![A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right] \\](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C)
now multiply the matrix with B
![X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\](https://tex.z-dn.net/?f=X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2.83%5C%5C1%5C%5C4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
