We know that Angle Bisector Divides an Angle into Two Equal Angles.
As UP is the Angle Bisector of Angle U, It Divides Angle U into two Equal Parts they are Angle(1) and Angle(2)
⇒ Angle(1) = Angle(2)
Given Angle(1) = 5x + 10 and Angle(2) = 3x + 14
⇒ 5x + 10 = 3x + 14
⇒ 2x = 4
⇒ x = 2
⇒ Angle(1) = 5x + 10 = 5(2) + 10 = 10 + 10 = 20
So : Measure of Angle(1) is 20
Well, I bet you want your answer right away! So here it is.
<span>Given <span>f (x) = 3x + 2</span> and <span>g(x) = 4 – 5x</span>, find <span>(f + g)(x), (f – g)(x), (f × g)(x)</span>, and <span>(f / g)(x)</span>.</span>
To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.
(f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
<span>\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}<span><span>(<span><span>g</span><span>f</span><span></span></span>)</span>(x)=<span><span><span>g(x)</span></span><span><span>f(x)</span></span><span></span></span></span></span><span>= \small{\dfrac{3x+2}{4-5x}}<span>=<span><span><span>4−5x</span></span><span><span>3x+2</span></span><span></span></span></span></span>
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
<span>\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}<span><span>(<span><span>g</span><span>f</span><span></span></span>)</span>(x)=<span><span><span>4−5x</span></span><span><span>3x+2</span></span><span>
Hope I helped! :) If I did not help that's okay.
-Duolingo
</span></span></span></span>
Answer:
92% of Income is Rs 18400
Step-by-step explanation:
I = 18400/.92 = Rs 20000
First look for the fundamental solutions by solving the homogeneous version of the ODE:

The characteristic equation is

with roots
and
, giving the two solutions
and
.
For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

Assume the ansatz solution,



(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution
anyway.)
Substitute these into the ODE:




is already accounted for, so assume an ansatz of the form



Substitute into the ODE:





Assume an ansatz solution



Substitute into the ODE:



So, the general solution of the original ODE is
