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3 years ago

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For babysitting nicole charges a flat fee of $10 plus $5 per hour. She earned $27.50 on tuesday night. Write an equation to dete

rmine how many hours nicole babysat on tuesday night?

1 answer:

MrRissso [65]3 years ago

7 0

5h+10=$27.50, so h=3.5 hours

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Can someone help me please? <br> I’ll give you brainliest ^^

Advocard [28]

The sample space that represents all possible buses and boats combinations is given by:

{(D,1), (D,2), (D,3), (E,1), (E,2), (E,3), (F,1), (F,2), (F,3)}.

<h3>What is the sample space of an experiment?</h3>

The sample space is the <u>set that contains all possible outcomes for an experiment</u>.

In this problem, for each the bus and the boat, there are 3 outcomes, hence the cardinality of the sample space is given by:

3 x 3 = 9.

Considering that one letter from D, E and F, and one number for 1, 2 and 3 are chosen, the sample space is given by:

{(D,1), (D,2), (D,3), (E,1), (E,2), (E,3), (F,1), (F,2), (F,3)}.

More can be learned about sample spaces at brainly.com/question/25861936

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2 years ago

Plz Help!! <br><br> Factor the Polynomial: -6x² - 15x + 9

Salsk061 [2.6K]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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:

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

$A + B + E = 32$

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3 years ago

10. If f(x) = { 2x-5 if x < 4

julia-pushkina [17]

B is the answer to this question

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3 years ago

The population of a certain city increased by 22,000. Write a signed number to represent this temperature change

Ira Lisetskai [31]

There was no temperature change. Simple, easy math.

The POPULATION increased by 22000

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3 years ago