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3 years ago

The reciprocal function of sine is: A. cosine B. cosecant C. secant D. tangent

Mathematics

2 answers:

ollegr [7]3 years ago

8 0

Answer:

The reciprocal is cosecant that is option B

e-lub [12.9K]3 years ago

5 0

Answer:

B. cosecant

Step-by-step explanation:

Just took the test and got it wrong but this is the correct answer

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The range of the function f(x) = x + 5 is {7, 9). What is the function's domain?

Maru [420]

Answer:

(2,4)

Step-by-step explanation:

The range is the y values

y = x+5

Let y=7

7 =x+5

Subtract 5

7-5 x+5-5

2=x

Let y= 9

Subtract 5

9-5 x+5-5

4=x

The domain is (2,4)

8 0

3 years ago

The school is t blocks from Jim's house. The library is twice as far as the school is from Jim's house. How far is the library?

Degger [83]

Answer:

2t blocks

Step-by-step explanation:

We measure these distances from Jim's house.

The school is t blocks from Jim's house. Twice that is 2t blocks. Thus, the library is 2t blocks from Jim's house.

8 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$