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2 years ago

How to prove this using trig identities?

Mathematics

1 answer:

Pavlova-9 [17]2 years ago

7 0

Step-by-step explanation:

tanB + cotB = (sinB)/(cosB) + (cosB)/(sinB)

= (sin2B + cos2B)/[(cosB)(sinB)]

= 1/[(cosB)(sinB)]

= (1/cosB)(1/sinB)

= (secB)(cscB)

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Find the equation that has a slope of - 1/8 and passes through the point (0.-4)

nikdorinn [45]

Answer:

ez, answer choice number 4

Step-by-step explanation:

y=-1/8x(this is the slope) -4(this is the y intercept

4 0

2 years ago

Can I please have help me with this problem

guajiro [1.7K]

9514 1404 393

Answer:

B, C

Step-by-step explanation:

Shelby's total earnings will be the sum of the amount she has earned and the amount she can earn working h hours. That latter amount is 25h, since she makes $25 per hour. She wants that total to be at least $2000.

75 + 25h ≥ 2000

Dividing by 25 gives ...

3 +h ≥ 80

subtracting 3, we have ...

h ≥ 77

These match options B and C of your list.

8 0

2 years ago

Find the missing side lengths

Mashcka [7]

Answer:

y = 9√3

x = 18

Step-by-step explanation:

taking 30 as reference angle

b = y

p = 9

h = x

tan30 = p/b

1/√3 = 9/b

or, b = 9√3

$h^2 = p^2 + b^2\\x^2 = 9^2 + y^2\\or, x^2 = 81 + (9\sqrt{3} )^2\\or, x^2 = 81 + 243\\or, x ^2 = 324\\or, x =\sqrt{324} \\\\so x = 18$

7 0

3 years ago

Solve for x. Help, please. ༼ つ ◕_◕ ༽つ

ahrayia [7]

Answer:

x = -10

Step-by-step explanation:

first you multiply both sides by -3

x + 7 = -3

then you subtract 7 from both sides

x = -10

hope this helps

5 0

2 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

2 years ago