Answer:
I would say it would also be 64 degrees because if you flip the HGJ to the IJG it would be the same! Hope this helps
:)

as you notice above, is the first-row components from A, multiplying all the columns subsequently on B, and you add the products of that row, that gives you one component on the AB matrix
in the one above, we end up with a 2x3 AB matrix
hello
the question here is
![3\sqrt[]{7}(14-4\sqrt[]{56})](https://tex.z-dn.net/?f=3%5Csqrt%5B%5D%7B7%7D%2814-4%5Csqrt%5B%5D%7B56%7D%29)
step 1
multiply through the bracket by the coeffiecient
![\begin{gathered} 3\sqrt[]{7}(14-4\sqrt[]{56}) \\ (3\sqrt[]{7}\times14)-(3\sqrt[]{7}\times4\sqrt[]{56}) \\ (14\times3\sqrt[]{7})-3\sqrt[]{7}\times4\sqrt[]{4\times14} \\ (42\sqrt[]{7})-3\sqrt[]{7}\times8\sqrt[]{14} \\ (42\sqrt[]{7})-\lbrack(3\times8)\sqrt[]{7\times14} \\ 42\sqrt[]{7}-24\sqrt[]{98} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%203%5Csqrt%5B%5D%7B7%7D%2814-4%5Csqrt%5B%5D%7B56%7D%29%20%5C%5C%20%283%5Csqrt%5B%5D%7B7%7D%5Ctimes14%29-%283%5Csqrt%5B%5D%7B7%7D%5Ctimes4%5Csqrt%5B%5D%7B56%7D%29%20%5C%5C%20%2814%5Ctimes3%5Csqrt%5B%5D%7B7%7D%29-3%5Csqrt%5B%5D%7B7%7D%5Ctimes4%5Csqrt%5B%5D%7B4%5Ctimes14%7D%20%5C%5C%20%2842%5Csqrt%5B%5D%7B7%7D%29-3%5Csqrt%5B%5D%7B7%7D%5Ctimes8%5Csqrt%5B%5D%7B14%7D%20%5C%5C%20%2842%5Csqrt%5B%5D%7B7%7D%29-%5Clbrack%283%5Ctimes8%29%5Csqrt%5B%5D%7B7%5Ctimes14%7D%20%5C%5C%2042%5Csqrt%5B%5D%7B7%7D-24%5Csqrt%5B%5D%7B98%7D%20%5Cend%7Bgathered%7D)
Answer:
Step-by-step exExplanation:
4
x
=
7
4
−
2
x
=
1
4
2
x
=
1
(
4
x
)
2
=
1
(
7
)
2
=
49
Answer:
x = 2 or x = -2
Step-by-step explanation:
7x^2 - 28 = 0
7x^2 = 28
x^2 = 28/7 = 4
x = root(4) or x = -root(4)
x = 2 or x = -2