Answer:
3 x 34 = 102 + 2 = 104 is your answer just do the equation
Answer:
74.0°
Step-by-step explanation:
In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the nearest 10th of a degree
Solution:
A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.
Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.
Using sine rule, we can find ∠K:

Hello,
log base 9(m/(m-4))=-2
==>m/(m-4)=9^(-2)
==>81m=m-4
==>80m=-4
==>m=-1/20
For the first 60 positive integers, a = 1, n = 60, l = 60.
Sn = n/2(a + l)
s = 60/2(1 + 60) = 30(61)
For the next 60 positive integer, a = 61, n = 60, l = 120
Sum = 60/2(61 + 120) = 30(61 + 120) = 30(61) + 30(120) = s + 3600
Sum of first 120 positive integers = s + s + 3600 = 2s + 3600
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)