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3 years ago

If P (x,y) is the point on the unit circle defined by real number 0, then csc0=_.

Mathematics

1 answer:

dezoksy [38]3 years ago

7 0

Answer:

cscθ = $\frac{1}{y}$

Step-by-step explanation:

cscθ = $\frac{Hypotenuse}{opposite}$

For a unit circle, radius = 1

hence, hypotenuse = 1

opposite = y

cscθ = $\frac{1}{y}$

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Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

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

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

6 0

3 years ago

Estimate by rounding the largest number to hundreds and then multiplying.321 x 5 is approximately

Zanzabum

Answer:

1600

Step-by-step explanation:

1600 is the answer because

321×5 = 1605 so estimation = 1605

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3 years ago

The point (1,-1) is on the terminal side of angle theta in standard position what are the values of sine cuisine and tangent of

pickupchik [31]

$\bf \textit{terminal point }\implies 
\begin{array}{lclcll}
(&1&,&-1)\\
&\uparrow &&\uparrow \\
&x&&y
\end{array}
\\\\\\
sin(\theta)=\cfrac{y}{r}
\qquad 
% cosine
cos(\theta)=\cfrac{x}{r}
\qquad 
% tangent
tan(\theta)=\cfrac{y}{x}\\\\
-----------------------------\\\\
\textit{so, what's the radius "r"?}\implies r^2=x^2+y^2\implies r=\sqrt{x^2+y^2}$

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Answer: income

Step-by-step explanation:

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3 years ago

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katovenus [111]

Answer:

1 2/7

Step-by-step explanation:

4.5:3 1/2

45/10 ÷ 7/2

45/10 x 2/7

90/70=9/7

=1 2/7

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3 years ago

Read 2 more answers

If P (x,y) is the point on the unit circle defined by real number 0, then csc0=_.​

If P (x,y) is the point on the unit circle defined by real number 0, then csc0=_.