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3 years ago

A sequence is defined by the recursive function f(n+1)=f(n)-2 If f(1)=10 what is f(3)?

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

4 0

Given :- 

A sequence is defined by f(n + 1) = f(n) -2 .
f(1) = 10

To Find :- 

The value of f(3) .

Solution :- 

As per Question ,

→ f(n + 1) = f(n) - 2

→ f(n + 1 ) - f(n ) = -2

Therefore , the second term of the sequence is ,

→ f(2) - f(1) = -2

→ f(2) - 10 = -2

→ f(2) = 10 -2

→ f(2) = 8

Similarly,

→ f(3) - f(2) = -2

→ f(3) = 8 -2

→ f(3) = 6

<h3>Hence the value of f(3) is 6 .</h3>

DENIUS [597]3 years ago

3 0

Answer:

f(3) = 6

Step-by-step explanation:

Using the recursive rule and f(1) = 10 , then

f(2) = f(1) - 2 = 10 - 2 = 8

f(3) = f(2) - 2 = 8 - 2 = 6

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Is 8/10 mile longer than 1/2

seropon [69]

Yes because if you find common denominators (in this case, "10") you would have to compare 8/10 to 5/10. 8>5 so 8/10 is longer than 1/2.

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3 years ago

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What values of a and b make the equation true? StartRoot 648 EndRoot = StartRoot 2 Superscript a Baseline times 3 Superscript b

kramer

Option C: $a=3, b=4$ is the value of a and b

Explanation:

Given that the expression $\sqrt{648}=\sqrt{2^a\times3^b}$

We need to determine the value of a and b

Let us consider the term $\sqrt{648}$ and take the prime factorization of the term 648

Thus, we have,

648 divides by 2,

$648=2 \cdot 324$

324 divides by 2,

$648=2 \cdot 2 \cdot 162$

162 divides by 2,

$648=2 \cdot 2 \cdot 2 \cdot 81$

81 divides by 3,

$648=2 \cdot 2 \cdot 2 \cdot 3 \cdot 27$

27 divides by 3,

$648=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 9$

9 divides by 3,

$648=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3$

Thus, we have,

$\sqrt{648}=\sqrt{2^{3} \cdot 3^{4}}$

Therefore, equating the powers of 2 and 3, we get,

$a=3, b=4$

Hence, the value of a and b is 3 and 4

Thus, Option C is the correct answer.

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3 years ago

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Part A What are the possible outcomes of rolling a six-sided die?

nikdorinn [45]

Answer:

the possible outcomes are :

one (1)
two (2)
three (3)
four (4)
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3 years ago

Please help me I'll give brainliest

Olenka [21]

Hey, there isn't a question, but I'd love to help!

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3 years ago

Using Rolle's theorem prove that the function has at most one root on the given interval <img src="https://tex.z-dn.net/?f=f%28x

Sophie [7]

Answer:

Show that if $f(a) = f(b) = 0$ for some $a,\, b \in [-2,\, 2]$ where $a \ne b$ , then by Rolle's Theorem $f^{\prime}(x) = 0$ for some $x \in (-2,\, 2)$ . However, no such $x$ exists since $f^{\prime}(x) < 0$ for all $x \in (-2,\, 2)\!$ .

Note that Rolle's Theorem alone does not give the exact value of the root. Neither does this theorem guarantee that a root exists in this interval.

Step-by-step explanation:

The function $f(x) = x^{3} - 12\, x + 11$ is continuous and differentiable over $[-2,\, 2]$ . By Rolle's Theorem. if $f(a) = f(b)$ for some $a,\, b \in [2,\, -2]$ where $a \ne b$ , then there would exist $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$ .

Assume by contradiction $f(x)$ does have more than one roots over $[-2,\, 2]$ . Let $a$ and $b$ be (two of the) roots, such that $a \ne b$ . Notice that $f(a) = 0 = f(b)$ just as Rolle's Theorem requires. Thus- by Rolle's Theorem- there would exist $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$ .

However, no such $x \in (a,\, b)$ could exist. Notice that $f^{\prime}(x) = 3\, x^{2} - 12$ , which is a parabola opening upwards. The only zeros of $f^{\prime}(x)$ are $x = (-2)$ and $x = 2$ .

However, neither $x = (-2)$ nor $x = 2$ are included in the open interval $(-2,\, 2)$ . Additionally, $a,\, b \in [-2,\, 2]$ , meaning that $(a,\, b)$ is a subset of the open interval $(-2,\, 2)$ . Thus, neither zero would be in the subset $(a,\, b)$ . In other words, there is no $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$ . Contradiction.

Hence, $f(x) = x^{3} - 12\, x + 11$ has at most one root over the interval $[-2,\, 2]$ .

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2 years ago