10[(6 + 4) / 2]
10[10/2]
10 * 5
50 <==
We can solve for the length of side a to the nearest whole number using the Laws of Cosines such as the formula is shown below:
a²=b²+c²-2bcCosA
Solving for the value of a, we have:
a²=10²+14²-2(10)(14)cos54°
a²=131.42
a=11.46
The answer is 11.46 or 11.5.
Gggggggggggggggggggggggggggggggggggggg. It would be 7
Solve for x:
x^2 + 4 x + 25 = 0 I ssume that's the notation.
Subtract 25 from both sides:
x^2 + 4 x = -25
Add 4 to both sides:
x^2 + 4 x + 4 = -21
Write the left hand side as a square:
(x + 2)^2 = -21
Take the square root of both sides:
x + 2 = i sqrt(21) or x + 2 = -i sqrt(21)
Subtract 2 from both sides:
x = i sqrt(21) - 2 or x + 2 = -i sqrt(21)
Subtract 2 from both sides:
Answer: x = i sqrt(21) - 2 or x = -i sqrt(21) - 2