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salantis [7]
3 years ago
13

Giselle pays $190 in advance on her account at the athletic club. Each time she uses the club, $10 is deducted from the account.

Model the situation with a linear function.
Mathematics
2 answers:
ch4aika [34]3 years ago
7 0
It has to be 19 if not tell answer choices please so i can help you

snow_tiger [21]3 years ago
3 0

x=190-10(n) im pretty sure this is correct because 10 times n would be the amount of times went to the place and it would be subtracted from 190


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If the value of X is negative what equation still be true ?explain.
Brut [27]

Answer:

yes

Step-by-step explanation:

5 0
3 years ago
For a segment of a radio show, a disc jockey can play 10 records. there are 14 records to select from, in how many ways can the
mixas84 [53]
The formula of combination is 

nCr= \frac{n!}{(n-r)!r!}

We have n=14 and r=10

14C10= \frac{14!}{(14-10)!10!} = 1001

Hence, there are 1001 ways of arranging 10 records from 14 records 
3 0
3 years ago
A company purchased $10,000 of merchandise on January 5 with terms 2/10, n/30. On January 7, it returned $1,200 worth of merchan
s2008m [1.1K]

Answer:

C. Debit Accounts Payable $8,800; credit Merchandise Inventory, $176; credit Cash $8,624.

Step-by-step explanation:

Data given in the question is inconsistent with the options given.

Terms 2/10, n/30 means there is a discount of 2% is available on payment of due amount within discount period of 10 days after sale with net credit period of 30 days.

Purchases = $10,000

Returns = $1,200

Amount Due = $10,000 - $1,200 = $8,800

As the payment is made after discount period, so no discount will be availed. Full amount of $8,800 will be paid.

A similar and correct question is given below and answer is made accordingly.

A company purchased $10,000 of merchandise on January 5 with terms 2/10, n/30. On January 7, it returned $1,200 worth of merchandise. On January 12, it paid the full amount due. Assuming the company uses a perpetual inventory system, and records purchases using the gross method, the correct journal entry to record the payment on January 12 is:

Debit Accounts Payable $10,000; credit Merchandise Inventory $200; credit Cash $9,800.

Debit Merchandise Inventory $8,800; credit Cash $8,800.

Debit Accounts Payable $8,800; credit Merchandise Inventory, $176; credit Cash $8,624.

Debit Cash $1,600; credit Accounts Payable $1,600.

Debit Accounts Payable $8,624; credit Cash $8,624.

Solution

Terms 2/10, n/30 means there is a discount of 2% is available on payment of due amount within discount period of 10 days after sale with net credit period of 30 days.

Purchases = $10,000

Returns = $1,200

Amount Due = $10,000 - $1,200 = $8,800

As the payment is made within discount period, so discount will be availed

Discount = $8,800 x 2% = $176

Cash Paid = $8,800 - $176 = $8,624

5 0
3 years ago
Write a polynomial function in standard form with zeros at 0, 1, and 2.
BARSIC [14]

Answer:

Step-by-step explanation:

f(x)=(x-0)(x-1)(x-2)

=x(x^2-3x+2)

or f(x)=x^3-3x^2+2x

3 0
3 years ago
High school math. Please answer everything in picture. Thank you................................................................
vaieri [72.5K]

Answer:

3) Cubic polynomial with four terms.

4) Linear polynomial with two terms.

5) The zeros are: (1-sqrt(5))/2=-0.618, 1, and (1+sqrt(5))/2=1.618

The y-intercept is y=1

Please, see the attached graph.

6) The zeros are: -1.272, 1.272

The y-intercept is y=1

Please, see the attached graph.

7) Please, see the attached files.

8) Please, see the attached files.

9) Please, see the attached files.

10) Please, see the attached files.


Step-by-step explanation:

3) Degree is the maximun exponent that the variable "x" has, in this case 3, then this is a cubic polynomial, and it has four terms (6x^3, -7x^2, -10x, and -8).


4) Degree is the maximun exponent that the variable "x" has, in this case 1 (x=x^1), then this is a linear polynomial, and it has two terms (-10x, and 10).


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

Zeros:

f(x)=0→x^3-2x^2+1=0

Factoring:

    x^3-2x^2+0x+1  

 ! 1      -2       0    1

<u>1 !____1____-1_-1</u>

   1      -1        -1   0

   1x^2-1x      -1 = x^2-x-1

x^3-2x^2+1=0→(x-1)(x^2-x-1)=0

The zeros are:

x-1=0→x-1+1=0+1→x1=1

x^2-x-1=0

ax^2+bx+c=0; a=1, b=-1, c=-1

x=(-b+-sqrt(b^2-4ac))/(2a)

x=(-(-1)+-sqrt((-1)^2-4(1)(-1))/(2(1))

x=(1+-sqrt(1+4))/2

x2=(1+-sqrt(5))/2=(1+2.236067978)/2=3.236067978/2=1.618033989=1.618

x=(1-sqrt(5))/2=(1-2.236067978)/2=-1.236067978/2=-0.618033989=-0.618

x3=(1+sqrt(5))/2

The zeros are: (1-sqrt(5))/2=-0.618, 1, and (1+sqrt(5))/2=1.618  

The y-intercept is:

x=0→f(0)=(0)^3-2(0)^2+1→f(0)=0-2(0)+1→f(0)=0-0+1→f(0)=1

The y-intercept is y=1.


6) f(x)=-x^4+x^2+1

Zeros:

f(x)=0→-x^4+x^2+1=0

Factoring: Multiplyng both sides of the equation by -1:

(-1)(-x^4+x^2+1)=(-1)(0)

x^4-x^2-1=0

(x^2)^2-(x^2)-1=0

Changing x^2 by t:

t^2-t-1=0

at^2+bt+c=0; a=1, b=-1, c=-1

t=(-b+-sqrt(b^2-4ac))/(2a)

t=(-(-1)+-sqrt((-1)^2-4(1)(-1))/(2(1))

t=(1+-sqrt(1+4))/2

t1=(1+-sqrt(5))/2=(1+2.236067978)/2=3.236067978/2=1.618033989=1.618

t2=(1-sqrt(5))/2=(1-2.236067978)/2=-1.236067978/2=-0.618033989=-0.618

t^2-t+1=0→(t-1.618)(t-(-0.618))=0→(t-1.618)(t+0.618)=0

and t=x^2, then:

(x^2-1.618)(x^2+0.618)=0

Factoring the first parentheses using a^2-b^2=(a+b)(a-b), with:

a^2=x^2→sqrt(a^2)=sqrt(x^2)→a=x

b^2=1.618→sqrt(b^2)=sqrt(1.618)→b=1.272

(x^2-1.618)(x^2+0.618)=0→(x+1.272)(x-1.272)(x^2+0.618)=0

The zeros are:

x+1.272=0→x+1.272-1.272=0-1.272→x1=-1.272

x-1.272=0→x-1.272+1.272=0+1.272→x2=1.272

The zeros are: -1.272, and 1.272  

The y-intercept is:

x=0→f(0)=-(0)^4+(0)^2+1→f(0)=-0+0+1→f(0)=1

The y-intercept is y=1.

3 0
3 years ago
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