Verify please dont understand
Answer:
Statement 3
Step-by-step explanation:
∠GAF ≅ ∠TAE
Since, there is no relation between these angles,
Therefore, statement is False.
∠TAE and ∠FAH are complementary.
If the sum of the two angles is 90°, pair of angles is said to be complimentary angles.
Therefore, this statement is False.
∠TAF is obtuse.
Measure of obtuse angle is more than 90°
From the given picture, measure of ∠TAF = m∠TAH + m∠HAF
= 90° + 50°
= 140°
Here, m∠TAF > 90°
Therefore, ∠TAF is an obtuse angle.
Statement (3) is the correct statement.
The formula for illuminance is given by
E = I / d^2
This formula only holds true for one-dimensional illuminance
The problem asks for the illuminance across the floor. We need to use two variables, x and y.
From Pythagorean Theorem
d^2 = x^2 + y^2
and from Trigonometry
x = d cos t
y = d sin t
The function for the illuminance can be represented by the composite function
E = I cos² t / x²
and
E = I sin² t / y²
The boundary of these functions is:
<span>0 < t < 8
So, the value of t must be in radians and not in degrees</span>
Answer:
x = - 5, x = - 3
Step-by-step explanation:
Given
x² + 15 = - 8x ( add 8x to both sides )
x² + 8x + 15 = 0 ← in standard form
Consider the factors of the constant term (+ 15) which sum to give the coefficient of the x- term (+ 8)
The factors are 5 and 3, since
5 × 3 = 15 and 5 + 3= 8, thus
(x + 5)(x + 3) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 5 = 0 ⇒ x = - 5
x + 3 = 0 ⇒ x = - 3
Step-by-step explanation:
<h3>Appropriate Question :-</h3>
Find the limit
![\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%5E2-x%7D-%5Cdfrac%7B1%7D%7Bx%5E3-3x%5E2%2B2x%7D%5Cright%5D)
Given expression is
On substituting directly x = 1, we get,
which is indeterminant form.
Consider again,
can be rewritten as
![\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28%20%7Bx%7D%5E%7B2%7D%20-%203x%20%2B%202%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28%20%7Bx%7D%5E%7B2%7D%20-%202x%20-%20x%20%2B%202%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28%20x%28x%20-%202%29%20-%201%28x%20-%202%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%7B%28x%20-%202%29%7D%5E%7B2%7D%20-%201%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%28x%20-%202%20-%201%29%28x%20-%202%20%2B%201%29%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%28x%20-%203%29%28x%20-%201%29%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%28x%20-%203%29%7D%7Bx%28x%20-%202%29%7D%5Cright%5D)
Hence,
![\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}](https://tex.z-dn.net/?f=%5Crm%5Cimplies%20%5C%3A%5Cboxed%7B%20%5Crm%7B%20%5C%3A%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%5E2-x%7D-%5Cdfrac%7B1%7D%7Bx%5E3-3x%5E2%2B2x%7D%5Cright%5D%20%3D%202%20%5C%3A%20%7D%7D)
