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3 years ago

Find the equation of the line passing through the point (2,-4) that is parallel to the line y=3x+2

Mathematics

1 answer:

m_a_m_a [10]3 years ago

7 0

Answer:

y=3x-10

Step-by-step explanation:

When you have to find a parallel line, you take the slope of the other line, so we already know y=3x+b. Then, plug in 2 and -4 as your x and y values and you get -4=3(2)+b. Then solve 3×2=6, so -4=6+b, subtract 6 from both sides, -10=b.

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Two thirds equals 18 over x plus 5

Wewaii [24]

Problem: 2/3= 18/(x+5)

First, I would multiply the (x+5) over

Making it 2/3(x+5)=18

Second, I would Divide by 2/3 (which is the same as multiplying by the reciprocal 3/2)

which shows up like this now (x+5)=18*3/2

third I would subtract the 5 over

which will give you your answer

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3 years ago

Camilla makes and sells jewelry. she has 8,160 silver beads and 2,880 black beads to make necklaces. Each necklace will contain

liubo4ka [24]

8160 dived by 85 equals 96, 2880 divided by 30 equals 96, therefore she can make 96 necklaces

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3 years ago

Read 2 more answers

Write the equation of the graph

olchik [2.2K]

Answer:

$y = 2.cos(x) - 2$

Step-by-step explanation:

1. Shifting $cos(x)$ right or left, top or down, as the graph in the question has the same illustration of $cos(x)$ . Consider

$y = a.cos(x) + b$

2. find a and b by using two sample points from the graph: $(0, 0)$ and $(\pi, -4)$

$0 = a.cos(0) + b => a + b = 0\\ -4 = a.cos(\pi) + b => b-a=-4\\\left \{ {{2b=-4} \atop {2a=4}} \right.\\a = 2, b=-2\\$

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3 years ago

Is parallel to 3x-5y=7 and passes through (0,-6)

oksian1 [2.3K]

Yes yes it isssssssssss

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3 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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3 years ago