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ANTONII [103]
1 year ago
10

Given ​ f(x)=x^2+12x+26 ​.

Mathematics
1 answer:
lapo4ka [179]1 year ago
6 0

f(x)=x^2+12x+26

Vertex form of f(x) = a(x-h)^2 +k

We change f(x) into vertex form using completing the square method

f(x)=x^2+12x+26

We take half of square of middle term

middle term is 12

Half of square of 12 is \frac{12}{2}=6 then 6^2 = 36

Add and subtract 36 to complete the square

f(x)=(x^2+12x+36)+26-36

f(x)= (x+6)^2 +26 - 36 f(x)= (x+6)^2 -10

The vertex form f(x)= (x+6)^2 -10

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11 hours

Step-by-step explanation:

27 miles per hour. 297/27 equals 11. to double check 11 times 27 equals 297.

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Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in
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Answer with step-by-step explanation:

We are given that the recurrence relation

f_n=5f_{n-4}+3f_{n-5}

for n=5,6,7,..

Initial condition

f_0=0,f_1=1,f_2=1,f_3=2,f_4=3

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

f_n=f_{n-1}+f_{n-2},f_0=0,f_1=1

Apply this

f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}

f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}

f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}

Substitute n=2

f_2=f_1+f_0=1+0=1

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Now, we have to show that f_{5n} is divisible by 5 for n=1,2,3,..

Now replace n by 5n

f_{5n}=5f_{5n-4}+3f_{5n-5}

Apply induction

Substitute n=1

f_5=5f_1+3f_0=5+0=5

It is true for n=1

Suppose it is true for n=k

f_{5k}=5f_{5k-4}+3f_{5k-5} is divisible 5

Let f_{5k}=5q

Now, we shall prove that for n=k+1 is true

f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)

f_{5k+5}=5(f_{5k+1}+3q)

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the f_{5n} is divisible by 5 for n=1,2,3,..

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