2. Which system of inequalities are represented by the graph below:

What’s the quotient of the 1/4 divided by 1/16

dusya [7]

1/4 divided by 1/16 is the same thing as 1/4 times 16/1 so that would multiple to 16/4 which is 4.

3 0

3 years ago

Joel earns $1,700 per month. If he spends $425 on rent each month, what percent of his income does he spend on rent?

a_sh-v [17]

The answer is 25%
you take (425/1700)*100
0.25 * 100% = 25%

4 0

3 years ago

Express the terms of the following sequence by giving a recursive formula.

emmasim [6.3K]

The recursive formula for given sequence is: $a_n = a_{n-1}-7$

And the terms will be expressed as:

$a_1 = 11\\a_2 = a_{2-1} - 7 \\a_2= a_1 - 7\\a_2 = 11- 7\\a_2 = 4\\a_3 = a_{3-1} - 7 \\a_3= a_2 - 7\\a_3 = 4 - 7\\a_3 = -3\\a_4 = a_{4-1} - 7 \\a_4= a_3 - 7\\ a_4= -3 - 7\\a_4 = -10\\a_5 = a_{5-1} - 7 \\a_5= a_4 - 7\\a_5 = -10 - 7\\a_5 = -17\\$

Step-by-step explanation:

First of all, we have to determine if the given sequence is arithmetic sequence or geometric. For that purpose, we calculate the common difference and common ratio

Given sequence is:

11,4,-3,-10,-17...

Here

$a_1 = 11\\a_2 = 4\\a_3 = -3\\So,\\d = a_2 - a_1 = 4-11 = -7\\d = a_3-a_2 = -3-4 = -7$

As the common difference is same, given sequence is an arithmetic sequence.

A recursive formula is a formula that is used to generate the next term of the sequence using the previous term and common difference

So, the recursive formula for an arithmetic sequence is given by:

$a_n = a_{n-1} +d\\Putting\ d = -7\\a_n = a_{n-1}-7$

Hence,

The recursive formula for given sequence is: $a_n = a_{n-1}-7$

And the terms will be expressed as:

$a_1 = 11\\a_2 = a_{2-1} - 7 \\a_2= a_1 - 7\\a_2 = 11- 7\\a_2 = 4\\a_3 = a_{3-1} - 7 \\a_3= a_2 - 7\\a_3 = 4 - 7\\a_3 = -3\\a_4 = a_{4-1} - 7 \\a_4= a_3 - 7\\ a_4= -3 - 7\\a_4 = -10\\a_5 = a_{5-1} - 7 \\a_5= a_4 - 7\\a_5 = -10 - 7\\a_5 = -17\\$

Keywords: arithmetic sequence, common difference

Learn more about arithmetic sequence at:

#LearnwithBrainly

6 0

4 years ago

Find the zeros (roots) of the following equations. f(x) = 2x5 - 9x4 + 12x3 - 12x2 + 10x - 3 = 0

Talja [164]

Hi,
$f(1)=2*1^5-9*1^4+12*1^3-12*1^2+10*1-3\\
=2-9+12-12+10-3\\
=0\\

f(x)=(x-1)(2x^4-7x^3+5x^2-7x+3)\\


g(x)=2x^4-7x^3+5x^2-7x+3\\

g(3)=2*3^4-7*3^3+5*3^2-7*3+3\\
=162-189+45-21+3\\
=0\\

g(x)=(x-3)(2x^3-x^2+2x-1)\\

h(x)=2x^3-x^2+2x-1\\

h( \dfrac{1}{2} )=2* \dfrac{1}{8} - \dfrac{1} {4}+1-1\\
=0\\

h(x)=(2x-1)(x^2+1)\\


f(x)=(x-1)(x-3)(2x-1)(x+i)(x-i)\\


$
Roots are 1,3,0.5,i,-i.

8 0

3 years ago

Olivia cashed a check

N76 [4]

Cool I guess hehehehehe

6 0

3 years ago