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3 years ago

What is the answer for this question please

Mathematics

2 answers:

Hunter-Best [27]3 years ago

4 0

The answer is 3 bc i used a book

Zigmanuir [339]3 years ago

3 0

Answer:

Step-by-step explanation:

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Determine the recursive function that defines the sequence.

Strike441 [17]

<h2>Answer:</h2>

Option: C is the correct answer.

C. $f(1)=4\\\\f(n)=5\cdot f(n-1)\ ;\ n\geq 2$

<h2>Step-by-step explanation:</h2>

Recursive Formula--

It is the formula which is used to represent the nth term of a sequence in terms of (n-1)th term of the sequence.

Here we are given a table of values by:

n f(n)

1 4

2 20

3 100

i.e. when n=1 we have:

$f(1)=4$

Also,

$f(2)=20\\\\i.e.\\\\f(2)=5\cdot 4\\\\i.e.\\\\f(2)=5\cdot f(1)$

Also,

$f(3)=100\\\\i.e.\\\\f(3)=5\cdot 20\\\\i.e.\\\\f(3)=5\cdot f(2)$

Hence, the recursive formula is:

$f(n)=5\cdot f(n-1)\ for\ n\geq 2$

5 0

3 years ago

The table below shows the heights of several books. Jean stacks a dictionary on top of her novel. how high is the stack of the t

tankabanditka [31]

Answer:between 5 and 5 1/2 inches

Step-by-step explanation:

8 0

3 years ago

Pls helpppppppppppp

Deffense [45]

-0.25 don’t know but I try

5 0

3 years ago

Given f(x)=17x-18, find x when f(x)=33

vichka [17]

33=17x-18 add 18 both sides
51=17x. divide 17 both sides
3 = x

3 0

3 years ago

Use logarithms to solve each equation 2x9/8=111

Illusion [34]

<span>We have this equation:
</span>
$2x^{\frac{9}{8}} = 111$

and we need to find the value of x.

First of all, we multiply the whole equation for 1/2, so our goal is to isolate x, therefore:

$x^{\frac{9}{8}}=\frac{111}{2}$

Next step we must do is to apply <span>logarithms:
</span>
$logx^{\frac{9}{8}} = log(\frac{111}{2})$

Next, we have to apply identities and then to solve the equation:

$\frac{9}{8}logx = log(\frac{111}{2})$
$logx = \frac{8}{9}log(\frac{111}{2})$
$logx = 1.5504$
$10^{logx} = 10^{1.5504}$

Finally, we have the value of x which was our goal. This is the answer for the question above:

$x= 35.5140$

7 0

3 years ago