<span><span><span>Given <span>f(x) = 2x + 3</span><span> and </span><span>g(x) = –x2 + 5</span><span>, find </span></span><span>(<span> f</span> </span>o<span> f )(x</span><span>).</span></span><span>(<span> f</span> </span>o<span> <span>f </span>)(x) = f (<span> f</span> (x))</span><span>
</span><span> = f (2x + 3)</span><span>
</span><span> = 2( ) + 3 </span>... setting up to insert the input<span>
</span><span> = 2(2x + 3) + 3</span><span>
</span><span> = 4x + 6 + 3</span><span>
</span> <span>= 4x + 9</span></span><span><span><span><span>Given </span><span>f(x) = 2x + 3</span><span> and </span><span>g(x) = –x2 + 5</span><span>, find </span></span><span>(g</span><span> o</span><span> g)(x</span><span>).</span></span><span>(g </span>o<span> g)(x) = g(g(x))</span><span>
</span><span> = –( )2 + 5 </span>... setting up to insert the input<span>
</span><span> = –(–x2 + 5)2 + 5</span><span>
</span><span> = –(x4 – 10x2 + 25) + 5</span><span>
</span><span> = –x4 + 10x2 – 25 + 5</span><span>
</span><span> = </span><span>–x4 + 10x2 – 20</span></span>
Sometimes you have to be careful with the domain and range of the composite function.
<span><span><span>Given </span><span>f (x) = sqrt(x)</span><span> and </span><span>g(x) = x – 2</span><span>, find the domains of </span><span>(<span> f</span> </span>o<span> g)(x)</span><span> and </span><span>(g </span>o<span> f )(x)</span>.</span></span>
( 0,0 ) is not the solution of the first inequality y≤x² +x-4 but ( 0,0) is the solution for the second inequality y <x²+2x+1.
<h3>What is inequality?</h3>
The relation between two expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than, or < ‘less than’
Finding the solution for the inequality is as follows:-
y ≤ x² +x-4 by putting x and y equal to 0.
0 ≤ 0 + 0 -4
0 ≤ - 4
This is incorrect so (0,0) can not be the solution for this inequality.
y < x²+2x+1.
0 < 0 + 0 + 1
0 < 1
This inequality is showing the solution for (0,0)
Therefore ( 0,0 ) is not the solution of the first inequality y≤x² +x-4 but ( 0,0) is the solution for the second inequality y <x²+2x+1.
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I'm not going to give you the answer, but I'm going to give really obvious hints to you.
Supplementary angles are when two angles add up to 180 degrees.
Complimentary is when two angles add up to 90 degrees.
I'm not sure how you find angles m<y and m<v, I can't even see the letter m on the attachment.
There are two angles that are adjacent to <GXF, and you are free to choose which ever ones you like: <FXE and <CXG
Answer:
35 books
Step-by-step explanation:
40*3=120
120-85=35
There are 5.1090942e+19 different ways this test can be completed
<h3>In how many different ways can this test be completed?</h3>
The given parameters are:
Questions, n =21
Options = 2 i.e. true or false
The number of different ways this test can be completed is calculated as:
Ways = n!
Substitute 21 for n in the above equation.
So, we have:
Ways = 21!
Expand the above equation
Ways = 21 * 20 * 19 * ...... 1
Evaluate the products
Ways = 5.1090942e+19
Hence, there are 5.1090942e+19 different ways this test can be completed
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