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igor_vitrenko [27]

3 years ago

14

Write 34/9 as a decimal. If necessary, use a bar to indicate which digit or group of digits repeats.

2 answers:

Elena L [17]3 years ago

7 0

Answer:

The decimal form of this fraction is 3.7 repeating.

rewona [7]3 years ago

5 0

Answer:

3.777...

Step-by-step explanation:

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Who can get the answer help me out

Alecsey [184]

I'd say it's D. I know the effect is that the money is increased, however I don't know by exactly how much.

8 0

4 years ago

If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

Nataly_w [17]

Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

-----------------------------------------------------------------------

Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

4 0

3 years ago

34=−2t can u help me

IrinaK [193]

34/2= -t => 17= -t => t= -17

4 0

4 years ago

Taylor charges $15 for the first hour of dog walking and $10 for each additional hour or fraction of an hour. The rates that Luc

svetoff [14.1K]

Answer:

Taylor will charge more by $15

Step-by-step explanation:

Given

Taylor

First Hour = $15

Hourly Rate = $10

Lucy:

$f(x) = \left \{ {{20x,\ 0 \le \ x \ \le 2} \atop {10x, x>2}} \right.$

For a dog walk of 2.5 hours.

Taylor's charges is:

$Charges = First\ Hour + Hourly\ Rate * 2.5$

$Charges = \$15 + \$10 * 2.5$

$Charges = \$15 + \$25$

$Charges = \$40$

Lucy's charges is:

Here, we consider

$f(x) = 10x$ Because $x > 2$

i.e 2.5 > 2

So, Lucy's charges is:

$Charges = 10 * \$2.5$

$Charges = \$25$

By comparison;

Taylor will charge more by $15

The difference is calculated by:

$Difference = \$40 - \$25$

$Difference = \$15$

4 0

3 years ago

What is the sum of the polynomials?

Allisa [31]

Answer:

$( - {x}^{2} + 9) + ( - 3 {x}^{2} - 11x + 4) \\ - {x}^{2} + 9 - 3{x}^{2} - 11x + 4 \\ = - 4 {x}^{2} - 11x + 13$

4 0

3 years ago

Read 2 more answers