I think it is D
I am not sure but this is what I think it is
Answer:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Step-by-step explanation:
So we have the expression:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
Recall the order of operations or PEMDAS:
P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.
E: Within the parentheses, if exponents are present, do them before all other operations.
M/D: Multiplication and division next, whichever comes first.
A/S: Addition and subtraction next, whichever comes first.
(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)
Thus, we should do the operations inside the parentheses first. Therefore:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
The parentheses is:

Square the 6 and the 4:

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

Therefore, the original equation is now:
![12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%5C%5C%3D12-%20%5B20-2%2848%29%5D)
Multiply with the brackets:
![=12-[20-96]](https://tex.z-dn.net/?f=%3D12-%5B20-96%5D)
Subtract with the brackets:
![=12-[-76]](https://tex.z-dn.net/?f=%3D12-%5B-76%5D)
Two negatives make a positive. Add:

Therefore:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Answer:
The answer is B.
Step-by-step explanation:
Given that the total angles in a triangle is 180° so in order to find the value of x, you have to substract 60° and 80° from 180° :




Answer:
HELLO DEAR,
GIVEN:-
10sin⁴A + 15cos⁴A = 6
=> 10(sin²A)² + 15cos⁴A = 6
=> 10{(1 - cos²A)²} + 15cos⁴A = 6
=> 10{1 + cos⁴A - 2cos²A} + 15cos⁴A = 6
=> 10 + 10cos⁴A - 20cos²A + 15cos⁴A = 6
=> 25cos⁴A - 20cos²A + 4 = 0
=> 25cos⁴A - 10cos²A - 10cos²A + 4 = 0
=> 5cos²A(5cos²A - 2) - 2(5cos²A - 2) = 0
=> (5cos²A - 2)(5cos²A - 2) = 0
=> cos²A = 2/5
=> cosA = √2/√5 [ secA = √5/√2]
therefore,
sinA = √[1 - 2/5] = √3/√5 [ cosecA = √5/√3]
now,
27cosec^6A + 8sec^6A
=> 27 × {(√5/√3)²}³ + 8 × {(√5/√2)²}³
=> 27 × 125/27 + 8 × 125/8
=> 125 + 125
=> 250.
HENCE, 27cosec^6A + 8sec^6A = 250.
I HOPE IT'S HELP YOU DEAR,
THANKS
Step-by-step explanation: