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3 years ago

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What are the answers too these questions

1 answer:

steposvetlana [31]3 years ago

8 0

Answer:

Step-by-step explanation:

i can't see it

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Reflections and glide reflections both give you a transformation where the preimage and image have opposite orientations.

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Three more than eight times a number is equal to 19

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8 meters in 10 seconds as a unit rate

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8 ÷ 10 = 0.8
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You can convert a temperature given in degrees Celsius to Fahrenheit temperature by using the expression 9x ÷ 5 + 32, where x is

Minchanka [31]

Water freezes at 32 degree Fahrenheit.

Step-by-step explanation:

Given,

Formula for Celsius to Fahrenheit = $\frac{9}{5}x+32$

x = Celsius temperature

Boiling point of water = 0°C

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3 years ago

Integrate the following problem:

vazorg [7]

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

$u=e^-^x$
$du=-e^-^x dx$

$dv=cos2x \ dx$
$v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$ .

Let's find u, du, dv, and v for this integral:

$u=-e^-^x$
$du=e^-^x dx$
$dv=\frac{sin2x}{2} dx$
$v=\frac{-cos2x}{4}$

Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$

Make the fractions have common denominators.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}$

Simplify this equation.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}$

Multiply the right side by the reciprocal of 5/4.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}$

Factor $e^-^x$ out of the numerator.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}$

Simplify this by using exponential properties.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}$

The final answer is $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C$ .

7 0

3 years ago

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