46 is between two integers. Name the integers and state the integer that is closest. plz i need help

Mathematics

1 answer:

Reika [66]3 years ago

3 0

Answer:

45 and 47

Step-by-step explanation:

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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:

Step-by-step explanation:

<u>Recursive Sequences</u>

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

Need help pls someone

dimaraw [331]

$\sqrt[3]{250a^{11}}=\sqrt[3]{125\cdot2a^{2+9}}=\sqrt[3]{125}\cdot\sqrt[3]{2a^2\cdot a^9}=5\sqrt[3]{2a^2}\cdot\sqrt[3]{a^9}\\\\=5\sqrt[3]{2a^2}\cdot\sqrt[3]{a^{3\cdot3}}=5\sqrt[3]{2a^2}\cdot\sqrt[3]{(a^3)^3}=5\sqrt[3]{2^2}\cdot a^3\\\\=\boxed{5a^3\sqrt[3]{2a^2}}$

7 0

3 years ago

A coyote had 10 pups in her third litter. It was twice as many as in her second litter, her second litter, and her second litter

Yuki888 [10]

A coyote had 10 pups in her third litter. (10)
It was twice as many as in her second litter,( 2( second litter) = third litter

her second litter had 3 more pups than her first.( first litter +3)=2nd litter

Find the number of pups in her litter