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3 years ago

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This penny will double every day for the next 30 days. What is the answer? Can someone tell me at least how to set up the doubli

ng equation for this doubling math excercise?

1 answer:

Law Incorporation [45]3 years ago

6 0

Answer:

10737418.24

Step-by-step explanation:

The answer is 2^n

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Someone help me with the ASAP!

JulsSmile [24]

$\mathsf{Given :\;\;\dfrac{12}{\frac{3}{5}}}$

$\mathsf{:\implies 12 \times \dfrac{5}{3}}$

$\mathsf{:\implies 4 \times 5}$

$\mathsf{:\implies 20}$

4 0

3 years ago

Read 2 more answers

The figure shows a circle inscribed in a triangle. To construct the inscribed circle, angle bisectors were first constructed at

klemol [59]

Answer:

Segments perpendicular to the sides of the triangle through the intersection of the angle bisectors were constructed.

Step-by-step explanation:

The above choice represents a bit of excess work. Actually, only one such perpendicular line segment needs to be constructed in order to determine the radius of the inscribed circle.

Once you know the center and radius, you can construct the inscribed circle.

7 0

3 years ago

Read 2 more answers

The rate on a cell phone is $0.22 per minute. If Izzy uses her phone for 2 hours, how much does she pay?

Molodets [167]

Answer:

$26.4

Step-by-step explanation:

ight 60 min in an hour

.22 per minute

.22 * 60 = 1 hour rate = $13.2

Multiply dat by 2 for 2 horas

13.2 * 2 = $26.4

5 0

3 years ago

specialty T-shirts are being sold online for $30 each plus a one-time handling fee of $2.75 . The total cost is a function of th

myrzilka [38]

30(5)+2.75=152.75

The answer is D.

30t +2.75; 152.75

5 0

3 years ago

The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th

Fantom [35]

Answer:

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C$

Step-by-step explanation:

To find:

∫ $6x^5(x^6-2)\,dx$

Solution:

Method of substitution:

Let $x^6-2=t$

Differentiate both sides with respect to $t$

$6x^5\,dx=dt$

[use $(x^n)'=nx^{n-1}$ ]

So,

∫ $6x^5(x^6-2)\,dx$ = ∫ $t\,dt$ = $\frac{t^2}{2}+C_1$ where $C_1$ is a variable.

(Use ∫ $t^n\,dt=\frac{t^{n+1} }{n+1}$ )

Put $t=x^6-2$

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1$

Use $(a-b)^2=a^2+b^2-2ab$

So,

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C$

where $C=2+C_1$

Without using substitution:

∫ $6x^5(x^6-2)\,dx$ = ∫ $6x^{11}-12x^5\,dx$ = $\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C$

So, same answer is obtained in both the cases.

7 0

2 years ago