lily has a $100 gift card to her favorite pastry shop.she spends $4 a day at the shop for the next 12 days

Mathematics

2 answers:

LiRa [457]4 years ago

6 0

Make a table X= each day y= amount of money X/y 0/100 1/96 2/92 3/89 4/85 5/81 6/76 7/72 8/68 9/64 10/60 11/56 12/52

katrin2010 [14]4 years ago

5 0

You would need to multiply $4 multiplied by the 12 day which will equal $48. Sence $48 tells the amount of money waisted, you need to subtract $100 - $48 and it will give you $52 .
So the answer is $52

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(1) [6pts] Let R be the relation {(0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0)} defined on the set {0, 1, 2, 3}. Find the foll

goldenfox [79]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

In point 1:

The Reflexive closure:

Relationship R reflexive closure becomes achieved with both the addition(a,a) to R Therefore, (a,a) is $(0,0),(1,1),(2,2) \ and \ (3,3)$

Thus, the reflexive closure: $R={(0,0),(0,1),(1,1),(1,2),(2,0),(2,2),(3,0), (3,3)}$

In point 2:

The Symmetric closure:

R relation symmetrically closes by adding(b,a) to R for each (a,b) of R Therefore, here (b,a) is: $(0,1),(0,2)\ and \ (0,3)$

Thus, the Symmetrical closure:

$R={(0,1),(0,2),(0,3)(1,0),(1,1)(1,2),(2,0),(2,2),(3,0), (3,3)}$

8 0

3 years ago

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.

marishachu [46]

Answer:

$\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

Step-by-step explanation:

We have been given the function

$f(x)=-2x^2+4x^3+3x^2+18$

From the rational zeros theorem, we have

$\text{Possible rational zeros}=\pm\frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}}$

From the given function,

Leading coefficient = 2

Factors of 2 are 1,2

Constant term = 18

Factors of constant term = 1, 2, 3, 6, 9, 18

Hence, we have

$\text{Possible rational zeros}=\pm\frac{1,2,3,6,9,18}{1,2}\\\\\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$