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2 years ago

Question is 100 points please help the question is in the pic

Mathematics

2 answers:

otez555 [7]2 years ago

6 0

For number 5: 180 -139 = x

For number 6: x = 41

For number 7: x + 18 = 90 (not 100% sure)

For number 8: x = 72

For number 9: you could write an equation about an acute angle and divide it into 2 parts the have one side 3 + 8 and have the other x. Then you could go from there. I’m sorry if you can’t figure out number 9. I tried to think of something.

mario62 [17]2 years ago

5 0

Answer:

For number 5: 180 -139 = x

For number 6: x = 41

For number 7: x + 18 = 90 (not 100% sure)

For number 8: x = 72

For number 9: you could write an equation about an acute angle and divide it into 2 parts the have one side 3 + 8 and have the other x. Then you could go from there. I’m sorry if you can’t figure out number 9. I tried to think of something.

Step-by-step explanation:

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Determine whether the lines are parallel, perpendicular, or neither. 5x + 2y = 14 and y = −5x + 9

valentinak56 [21]

Answer: Neither

Step-by-step explanation:

5x + 2y = 14

2y = -5x + 14

y = -5/2 x + 7; slope m = -5/2

y = –5x + 9 ; slope m = -5.

Parallel lines slopes are equal and perpendicular slopes are opposite and reciprocal.

So both lines are neither parallel nor perpendicular

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3 years ago

1. The speed of light is approximately 3 x 108 m/s. How far does light travel in 5.0 x 10 seconds?

leonid [27]

Answer:

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3 years ago

Read 2 more answers

Solve for x enter the solutions from the least to greatest x2+7=43 lesser x= Greater x=

Snowcat [4.5K]

Answer:

x = ±6

Step-by-step explanation:

Step 1: Write out equation

x² + 7 = 43

Step 2: Subtract 7 on both sides

x² = 36

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7 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:

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

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

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2 years ago