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4 years ago

Using the trigonometry tangent, find the values of Theta, Tan (Theta) = 6/16 *

Mathematics

1 answer:

Akimi4 [234]4 years ago

3 0

Answer:

Theta = 20.56 degrees

Step-by-step explanation:

Tan (theta) = 6/16.

Use the inverse of tan to find theta

Theta = $tan^{-1}$ (6/16)

Use your calculator to solve.

Theta = 20.56 degrees

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HELP BEEN HAVING A BAD DAY<br> Find the slope of each line (each block is one unit):

olchik [2.2K]

Answer:

m = -2/3

Step-by-step explanation:

m = slope

$m = \frac{y2 - y1}{x2 - x1}$

m = -3 -1/ 3 - (-3)

m= -4/ 6

m= -2/3

3 0

4 years ago

19. Which of the following

Andrew [12]

9514 1404 393

Answer:

b. angle K is acute

Step-by-step explanation:

We're often told not to draw any conclusions from the appearance of a figure in a geometry problem. Here, angle K appears to be somewhat less than 90°, so angle K is acute.

<em>Additional comment</em>

This choice of answer is confirmed by the fact that the other two (visible) choices say the same thing. If one of them is correct, so is the other one. Hence they must both be incorrect. (An obtuse angle is more than 90°.)

4 0

3 years ago

Round off 2409725 to nearest 10

Julli [10]

Answer:

Hey!

Step-by-step explanation:

Your answer is 2409725.0

Best of luck ♥

5 0

3 years ago

Read 2 more answers

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

Read 2 more answers

What is the answer for this 5-1/5x<5

Elanso [62]

The answer to this question is X>0 because if u work it out then the value of X will be greater than 0.

6 0

3 years ago