(tan(<em>x</em>) + cot(<em>x</em>)) / (tan(<em>x</em>) - cot(<em>x</em>)) = (tan²(<em>x</em>) + 1) / (tan²(<em>x</em>) - 1)
… = (sin²(<em>x</em>) + cos²(<em>x</em>)) / (sin²(<em>x</em>) - cos²(<em>x</em>))
… = -1/cos(2<em>x</em>)
Then as <em>x</em> approaches <em>π</em>/2, the limit is -1/cos(2•<em>π</em>/2) = -sec(<em>π</em>) = 1.
Answer:
The correct answer is:
(7s-2)+3+(s+3) = 52, or 8s+4 = 52.
Step-by-step explanation:
Since s is the son's age, "two less than seven times" the son's age would be represented by 7s-2. To represent this in 3 years, we would add 3: (7s-2)+3. In 3 years, the son's age, s, would be represented by s+3. We are told that the sum of these ages will be 52; this gives us (7s-2)+3+(s+3) = 52.
To simplify this, combine like terms. 7s+s = 8s; -2+3+3 = 4. This gives us 8s+4=52.
Answer: 6 if by whole number or 6.25 for decimals
Step-by-step explanation:
Answer:
18 and 9
OR
-18 and -9
Step-by-step explanation:
Let:
first number = x
second number = y
According to given conditions:
x - y = 9 ---------- eq1
x * y = 162 ------- eq2
From eq1:
x = 9 + y
Put it in eq2:
(9+ y)*y = 162
9y + y^2 - 162 = 0
OR
y^2 + 9y -162 = 0
By factorizing we get:
y^2 -9y + 18y - 162 = 0
Taking common:
y(y - 9) + 18(y - 9) = 0
(y - 9)(y + 18) = 0
So:
y = -18 and y = 9
ignoring negative integer
Putting values of y in eq1:
x - 9 = 9
x = 9 + 9
x = 18
So the two numbers are:
18 and 9
OR
-18 and -9
i hope it will help you!
