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pogonyaev
2 years ago
6

Math help please Let f(x)=x2+5 and g (x)=3x+4; find (g o f)(7); (g o f)(7)=; help me solve!! Please!

Mathematics
1 answer:
lorasvet [3.4K]2 years ago
5 0

f(x) = 2x

g(x) = x + 3

First let us find (fog)(x)

<span />

(fog)(x) = f(g(x)

             = f(x+3)

               = 2(x+3)

               = 2x + 6

==> (fog)(x) = 2x + 6

Now let us find (gof)(x):

(gof)(x) = g(f(x)

            = g(2x)

           = 2x + 3

==> <span>(gof)(x) = 2x + 3</span>

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Given the sequence 1/2 ; 4 ; 1/4 ; 7 ; 1/8 ; 10;.. calculate the sum of 50 terms
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<u>Hint </u><u>:</u><u>-</u>

  • Break the given sequence into two parts .
  • Notice the terms at gap of one term beginning from the first term .They are like \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8} . Next term is obtained by multiplying half to the previous term .
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<u>Solution</u><u> </u><u>:</u><u>-</u><u> </u>

We need to find out the sum of 50 terms of the given sequence . After splitting the given sequence ,

\implies S_1 = \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8} .

We can see that this is in <u>Geometric</u><u> </u><u>Progression </u> where 1/2 is the common ratio . Calculating the sum of 25 terms , we have ,

\implies S_1 = a\dfrac{1-r^n}{1-r} \\\\\implies S_1 = \dfrac{1}{2}\left[ \dfrac{1-\bigg(\dfrac{1}{2}\bigg)^{25}}{1-\dfrac{1}{2}}\right]

Notice the term \dfrac{1}{2^{25}} will be too small , so we can neglect it and take its approximation as 0 .

\implies S_1\approx \cancel{ \dfrac{1}{2} } \left[ \dfrac{1-0}{\cancel{\dfrac{1}{2} }}\right]

\\\implies \boxed{ S_1 \approx 1 }

\rule{200}2

Now the second sequence is in Arithmetic Progression , with common difference = 3 .

\implies S_2=\dfrac{n}{2}[2a + (n-1)d]

Substitute ,

\implies S_2=\dfrac{25}{2}[2(4) + (25-1)3] =\boxed{ 908}

Hence sum = 908 + 1 = 909

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Answer:

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Step-by-step explanation:

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y2 - y1 / x2 - x1

8 0
3 years ago
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riadik2000 [5.3K]

Answer:

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AURORKA [14]
Question: \frac{360}{4}
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Therefore the solution is 90. 
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