The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
Split up the integration interval into 4 subintervals:
![\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac%5Cpi8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi8%2C%5Cdfrac%5Cpi4%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi4%2C%5Cdfrac%7B3%5Cpi%7D8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%7B3%5Cpi%7D8%2C%5Cdfrac%5Cpi2%5Cright%5D)
The left and right endpoints of the
-th subinterval, respectively, are


for
, and the respective midpoints are

We approximate the (signed) area under the curve over each subinterval by

so that

We approximate the area for each subinterval by

so that

We first interpolate the integrand over each subinterval by a quadratic polynomial
, where

so that

It so happens that the integral of
reduces nicely to the form you're probably more familiar with,

Then the integral is approximately

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.
The answer is what is shown on the phone
Answer:
The final answers are x = 10.385 OR x = -0.385
Step-by-step explanation:
Given the equation is x^2 -4 = 10x
Rewriting it in quadratic form as:- x^2 -10x -4 = 0.
a = 1, b = -10, c = -4.
Using Quadratic formula as follows:- x = ( -b ± √(b² -4ac) ) / (2a)
x = ( 10 ± √(100 -4*1*-4) ) / (2*1)
x = ( 10 ± √(100 +16) ) / (2)
x = ( 10 ± √(116) ) / (2)
x = ( 10 ± 10.77 ) / (2)
x = ( 10 + 10.77 ) / (2) OR x = ( 10 - 10.77 ) / (2)
x = 20.77/2 OR x = -0.77/2
x = 10.385 OR x = -0.385
Hence, final answers are x = 10.385 OR x = -0.385