What is the equation of the tangent drawn for f(x)=x-x^2 which is parallel to the line 4x-y-3=0 pl help me °~°

y = 4x + 9/4

Step-by-step explanation:

4x − y − 3 = 0

y = 4x − 3

The slope of the line is 4.

The slope of the tangent line is:

f(x) = x − x²

f'(x) = 1 − 2x

Setting the slope equal to 4:

4 = 1 − 2x

2x = -3

x = -3/2

The point on f(x) is:

f(-3/2) = -3/2 − (-3/2)²

f(-3/2) = -3/2 − 9/4

f(-3/2) = -15/4

Using point-slope form:

y − (-15/4) = 4 (x − (-3/2))

y + 15/4 = 4 (x + 3/2)

y + 15/4 = 4x + 6

y = 4x + 9/4

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet

jeff jeff

step-by-step explanation: