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3 years ago

Find the range of the relation below, then determine whether the relation is a function.

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

4 0

Answer:

Step-by-step explanation:

It is a function because it passes the vertical line test and Xs do not repeat, then count out the Y-values which are the range.

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Find the value of ( 1+cosA) (1-cosA)

Mandarinka [93]

Answer:

sin^2 A

Step-by-step explanation:

first expand the trigonometry

1(1-cosA)+cosA(1-cosA)

=1-cosA+cosA-cos^2 A

= 1-cos^2 A

from trigonometric identity sin^2 A + cos^2 A= 1

sin^2 A= 1- cos^2 A

=sin^2 A

3 0

2 years ago

Find the value of x if 2x = 20 .

7nadin3 [17]

Answer:

x = 10

Step-by-step explanation:

Divide both sides by 2.

(2x) / 2 = 20 / 2

x = 10

8 0

3 years ago

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Hola feliz noche solo hago está pregunta para probar la aplicación cuántos es 7+1

butalik [34]

Answer:

es 8

Step-by-step explanation:

3 0

3 years ago

Jackets price is reduced from $25 to $15. by what percentage was the price of the sweater reduced.

Helga [31]

Answer:

40 percent

Step-by-step explanation:

10 / 25 =.4=40%

7 0

3 years ago

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If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

4 years ago