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Arisa [49]
3 years ago
11

Calculate a 15% Tip for $12.68

Mathematics
1 answer:
sleet_krkn [62]3 years ago
5 0
That would be $1.90 tip
You might be interested in
You have $5600. The best interest rate you can find is 3%
EleoNora [17]

Answer:

18 years

Step-by-step explanation:

The formula for computing accrued amount A for a principal of P at an interest rate of r(in decimal) compounded n times in a year for t years is given by

A = P(1 + \frac{r}{n})^{nt}

Note that r is percentage converted to decimal. So 3% = 3/100 = 0.03

We can rearrange the above equation to:

\frac{A}{P} = (1 + \frac{r}{n})^{nt}

Taking logs on both sides

log(\frac{A}{P}) = log(1 + \frac{r}{n})^{nt}

This gives

log(\frac{A}{P}) =nt \times log(1 + \frac{r}{n})\\So,\\nt = \frac{log(\frac{A}{P})}{ log(1 + \frac{r}{n})}

In this particular problem, n = 4, , A= 9600, P = 5600, r =0.03, so r/n = 0.03/4 = 0.0075

1 + r/n = 1+0.0075 = 1.0075

4t = log(9600/5600)/log(1.0075) = log(1.714) / log(1.0075) = 0.234 /0.00325 = 72

t = 72/4 = 18 years

4 0
2 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each verbal description of a sequen
galben [10]

Answer:

I think the question is wrong so, I will try and explain with some right questions

Step-by-step explanation:

We are give 6 sequences to analyse

1. an = 3 · (4)n - 1

2. an = 4 · (2)n - 1

3. an = 2 · (3)n - 1

4. an = 4 + 2(n - 1)

5. an = 2 + 3(n - 1)

6. an = 3 + 4(n - 1)

1. This is the correct sequence

an=3•(4)^(n-1)

If this is an

Let know an+1, the next term

an+1=3•(4)^(n+1-1)

an+1=3•(4)^n

There fore

Common ratio an+1/an

r= 3•(4)^n/3•(4)^n-1

r= (4)^(n-n+1)

r=4^1

r= 4, then the common ratio is 4

Then

First term is when n=1

an=3•(4)^(n-1)

a1=3•(4)^(1-1)

a1=3•(4)^0=3.4^0

a1=3

The first term is 3 and the common ratio is 4, it is a G.P

2. This is the correct sequence

an=4•(2)^(n-1)

Therefore, let find an+1

an+1=4•(2)^(n+1-1)

an+1= 4•2ⁿ

Common ratio=an+1/an

r=4•2ⁿ/4•(2)^(n-1)

r=2^(n-n+1)

r=2¹=2

Then the common ratio is 2,

The first term is when n =1

an=4•(2)^(n-1)

a1=4•(2)^(1-1)

a1=4•(2)^0

a1=4

It is geometric progression with first term 4 and common ratio 2.

3. This is the correct sequence

an=2•(3)^(n-1)

Therefore, let find an+1

an+1=2•(3)^(n+1-1)

an+1= 2•3ⁿ

Common ratio=an+1/an

r=2•3ⁿ/2•(3)^(n-1)

r=3^(n-n+1)

r=3¹=3

Then the common ratio is 3,

The first term is when n =1

an=2•(3)^(n-1)

a1=2•(3)^(1-1)

a1=2•(3)^0

a1=2

It is geometric progression with first term 2 and common ratio 3.

4. I think this correct sequence so we will use it.

an = 4 + 2(n - 1)

Let find an+1

an+1= 4+2(n+1-1)

an+1= 4+2n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=4+2n-(4+2(n-1))

d=4+2n-4-2(n-1)

d=4+2n-4-2n+2

d=2.

The common difference is 2

Now, the first term is when n=1

an=4+2(n-1)

a1=4+2(1-1)

a1=4+2(0)

a1=4

This is an arithmetic progression of common difference 2 and first term 4.

5. I think this correct sequence so we will use it.

an = 2 + 3(n - 1)

Let find an+1

an+1= 2+3(n+1-1)

an+1= 2+3n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=2+3n-(2+3(n-1))

d=2+3n-2-3(n-1)

d=2+3n-2-3n+3

d=3.

The common difference is 3

Now, the first term is when n=1

an=2+3(n-1)

a1=2+3(1-1)

a1=2+3(0)

a1=2

This is an arithmetic progression of common difference 3 and first term 2.

6. I think this correct sequence so we will use it.

an = 3 + 4(n - 1)

Let find an+1

an+1= 3+4(n+1-1)

an+1= 3+4n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=3+4n-(3+4(n-1))

d=3+4n-3-4(n-1)

d=3+4n-3-4n+4

d=4.

The common difference is 4

Now, the first term is when n=1

an=3+4(n-1)

a1=3+4(1-1)

a1=3+4(0)

a1=3

This is an arithmetic progression of common difference 4 and first term 3.

5 0
3 years ago
Select Is a Function or Is not a Function to correctly classify each relation
Gemiola [76]

Answer:

no

yes

no

yes

Step-by-step explanation:

x input repeats - 3

x input doesn't repeat

x input repeats -4

x input doesn't repeat

6 0
3 years ago
SOMEONE HELP ME WITH THIS ONE PLZ!
kolezko [41]
To construct a circle that circumscribes to a triangle, you would have to construct a circle that where all vertices of the triangle are on the circle. To do this you would have to construct the perpendicular bisectors of each side with your compass and straight edge. Comment on this answer if you are unsure of how to construct a perpendicular bisect (it's a long fundamental process to describe, and I wouldn't want to lecture you one something you already know). Once you have done so, set your compass point on the point where all perpendicular bisectors intersect (they should intersect in ONE point, if not you will have to redo it). Set your other compass lead on one of the vertices and spin away! If you have done this correctly, you should hit all three vertices when spinning your compass. Hope this helps!

Fun fact: the point where all perpendicular bisectors intersect is called the circumcenter
3 0
3 years ago
Which point is located at (2, -3)?point Hpoint Epoint Gpoint K
Fynjy0 [20]

Answer:

Since i dont have a graph il just tell you this way. on the x axis, go right 2 times, then go down 3 times and that will be your answer.

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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