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Vadim26 [7]
3 years ago
9

Jay on started off his penny collection with 1 penny. He then adds 5 pennies to his collection each day. How could you change th

e above scenario to make it a geometric series rather than an arithmetic series?
Mathematics
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

By doubling each day

Step-by-step explanation:

Jay is increasing  daily a fixed amount of pennies to his collection.  That's an arithmetic series, because to obtain the next term you have to do an addition.

To get a geometric series or progression you have to find the next term through multiplication.

So, if he wants to convert his series into a geometric series, he'd have to double his contribution to his collection each day (as an example).

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If 18√8 - 8 √18 = √n, what is n?
Yuliya22 [10]

Answer:

n=288

Step-by-step explanation:

Rewrite the equation as  

√

n

=

18

√

8

−

8

√

18

.

√

n

=

18

√

8

−

8

√

18

To remove the radical on the left side of the equation, square both sides of the equation.

√n

2

=

(

18

√

8

−

8

√

18

)

2

Simplify each side of the equation.  

Use  

n

√

a

x

=

a

x

n

to rewrite  

√

n  as  n

1

2

.

(

n

1

2

)

2

=

(

18

√

8

−

8

√

18

)

2

Simplify  

(

n

1

2

)

2

.  

Multiply the exponents in  

(

n

1

2

)

2

.  

Apply the power rule and multiply exponents,  

(

a

m)n

=

a

m

n

.

n

1

2

⋅

2

=

(

18

√

8

−

8

√

18

)

2

Cancel the common factor of  2  

Cancel the common factor.

n

1

2

⋅

2

=

(

18

√

8

−

8

√

18

)

2

Rewrite the expression.

n

1

=

(

18

√

8

−

8

√

18

)

2

Simplify.

n

=

(

18

√

8

−

8

√

18

)

2

Simplify  

(

18

√

8

−

8

√

18

)

2

Simplify each term.

Rewrite  

8  as  2

2

⋅

2

.  

Factor  

4  out of  8  

n

=

(

18

√

4

(

2

)

−

8

√

18

)

2

Rewrite  

4  as  2

2  

n

=

(

18√

2

2

2

−

8

√

18

)

2

Pull terms out from under the radical.

n

=

(

18

(

2

√

2

)

−

8

√

18

)

2

Multiply  

2  by  18  

n

=

(

36

√

2

−

8

√

18

)

2

Rewrite  

18

as  

3

2

⋅

2

.

Factor  

9

out of  

18

.

n

=

(

36

√

2

−

8

√

9

(

2

)

)

2

Rewrite  

9

as  

3

2

.

n

=

(

36

√

2

−

8

√

3

2

⋅

2

)

2

Pull terms out from under the radical.

n

=

(

36

√

2

−

8

(

3

√

2

)

)

2

Multiply  

3

by  

−

8

.

n

=

(

36

√

2

−

24

√

2

)

2

Simplify terms.

Subtract  

24

√

2

from  

36

√

2

.

n

=

(

12

√

2

)

2

Simplify the expression.

Apply the product rule to  

12

√

2

.

n

=

12

2

√

2

2

Raise  

12

to the power of  

2

.

n

=

144

√

2

2

Rewrite  

√

2

2

as  

2

.

Use  

n

√

a

x

=

a

x

n

to rewrite  

√

2

as  

2

1

2

.

n

=

144

(

2

1

2

)

2

Apply the power rule and multiply exponents,  

(

a

m

)

n

=

a

m

n

.

n

=

144

⋅

2

1

2

⋅

2

Combine  

1

2

and  

2

.

n

=

144

⋅

2

2

2

Cancel the common factor of  

2

.

Cancel the common factor.

n

=

144

⋅

2

2

2

Rewrite the expression.

n

=

144

⋅

2

1

Evaluate the exponent.

n

=

144

⋅

2

Multiply  

144

by  

2

.

n

=

288

5 0
3 years ago
30 POINTS
lesantik [10]

Answer: (26996, 42744)

Step-by-step explanation: N/A

5 0
3 years ago
What are some ways you can create a new system of equations based on a given system?
docker41 [41]
By adding/ subtraction systems
3 0
3 years ago
Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f
quester [9]

Answer:

It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x) (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

5 0
3 years ago
Read 2 more answers
What the approximate volume of the cylinder below in cubic centimeters
Molodets [167]

Answer:

V≈12308.8

Step-by-step explanation:

V=πr2h

V=(3.14)(14)2(20)

R=14cm

H=20cm

π=3.14

5 0
2 years ago
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