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bazaltina [42]
3 years ago
12

The Yoga class you wish to take is offering two price quotes: $108 for 5 weeks as (15 classes) or $125 for 8 weeks (24 classes).

How much per class would you save by taking the 8 week class?
A $2.10
B $2.05
C $1.99
D $1.89
Mathematics
2 answers:
WARRIOR [948]3 years ago
8 0

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Cost for 5 weeks i.e. 15 classes = $108

Cost of per class is given by

\frac{108}{15}=7.2

Cost for 8 weeks i.e. 24 classes = $125

Cost of per class is given by

\frac{125}{24}=5.21

Amount of saving of per class by taking the 8 week class is given by

7.2-5.21=\$1.99

Hence, Option 'C' is correct.

AURORKA [14]3 years ago
4 0
The answer is C)
108/15 - 125/24 = 1.99
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Which polynomial identity will prove that 19 = 27 − 8?
3241004551 [841]
The answer is Difference of Cubes.

Difference of cubes is a³ - b³.
27 can be written as 3³ (= 3 × 3 × 3 = 27). So, a = 3.
8 can be written as 2³ (= 2 × 2 × 2 = 8). So, b = 2.

Difference of cubes can be expressed as:
       a³ - b³ = (a - b)(a² + ab + b²)
⇒   3³ - 2³ = (3 - 2)(3² + 3×2 + 2²) = 1 × (9 + 6 + 4) = 1 × 19 = 19
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5 0
3 years ago
Evaluate using integration by parts ​
PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

I(n)=\displaystyle\int x^ne^x\,\mathrm dx

One round of IBP, setting

u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx

\mathrm dv=e^x\,\mathrm dx\implies v=e^x

gives

\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx

I(n)=x^ne^x-nI(n-1)

This is called a power-reduction formula. We could try solving for I(n) explicitly, but no need. n=5 is small enough to just expand I(5) as much as we need to.

I(5)=x^5e^x-5I(4)

I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)

I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)

I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)

I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))

Finally,

I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C

so we end up with

I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C

I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C

and the antiderivative is

\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C

8 0
3 years ago
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Answer:

8

Step-by-step explanation:

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