All categories

4 years ago

Plz help this is so hard

Mathematics

1 answer:

Mashutka [201]4 years ago

4 0

1) 5:4 2) 4:5 3) 5:9 4) 2:3 5) 1:4 6)3:2 7) 5:13 8) 6:7 9) 7:5 10) 13:18 11) 18:5 12) 6:13. The others were weird, because it's been forever since I've had to do these.

You might be interested in

A sequence can be generated by using fn = 2f(n-1)+1, where f1 = 4 and n is a whole number greater than 1. What are the first fou

hoa [83]

Answer:

{9,19,39,79}

Step-by-step explanation:

Recursive Sequences

The recursive sequence can be identified because each term is given as a function of one or more of the previous terms. Being n an integer greater than 1, then:

f(n) = 2f(n-1)+1

f(1) = 4

To find the first four terms of the sequence, we set n to the values {2,3,4,5}

For n=2

f(2) = 2f(1)+1

Since f(1)=4:

f(2) = 2*4+1

f(2) = 9

For n=3

f(3) = 2f(2)+1

Since f(2)=9:

f(3) = 2*9+1

f(3) = 19

For n=4

f(4) = 2f(3)+1

Since f(3)=19:

f(4) = 2*19+1

f(4) = 39

For n=5

f(5) = 2f(4)+1

Since f(4)=39:

f(5) = 2*39+1

f(5) = 79

4 0

3 years ago

Can anyone help me!

zhenek [66]

Answer:

$\large \text{$ \sf 2^{-64}$}$

Step-by-step explanation:

Given:

$\large \text{$ \sf (256)^{-4^{\frac{3}{2}}}$}$

This reads as: 256 to the power of "-4 to the power of 3/2".

Therefore, we need to deal with the "-4 to the power of 3/2" first.

Rewrite the 4 as 2²:

$\large \text{$ \sf \implies -(2^2)^{\frac{3}{2}}$}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies -2^{\left(2 \times \frac{3}{2}\right)}$}$

$\large \text{$ \sf \implies -2^{3}$}$

Therefore:

$\large \text{$ \sf \implies -2^3=(-2)(-2)(-2) = -8$}$

Replace "-4 to the power of 3/2" with -8 :

$\large \text{$ \sf \implies 256^{-8}$}$

Rewrite 256 as 2⁸ :

$\large \text{$ \sf \implies (2^8)^{-8}$}$

$\textsf{Again, apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies 2^{(8 \times -8)}$}$

$\large \text{$ \sf \implies 2^{-64}$}$

In one complete calculation:

$\large\begin{aligned} \sf (256)^{-4^{\frac{3}{2}}} & = \sf (256)^{-(2^2)^{\frac{3}{2}}}\\& = \sf (256)^{-2^{\left(2 \times \frac{3}{2}\right)}}\\& =\sf (256)^{-2^{3}}\\& = \sf (256)^{-8}\\& \sf = (2^8)^{-8}\\& = \sf 2^{(8 \times -8)}\\& =\sf 2^{-64}\end{aligned}$