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3 years ago

Plz help with this question

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

7 0

Answer:

y = -2x+4

Step-by-step explanation:

the equation of the line there is: y = -2x+8

The parallel line to this one should have the same slope as the first one(-2x)

Parallel lines always have the same slopes

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a sum of money is divided among two friends in the ratio 4:11 if the smaller amount is $420, determine the larger amount?

bixtya [17]

Answer:

1155

Step-by-step explanation:

let the ratio be 4x:11x

4x=420

x=105

11x=11*105

=1155

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2 years ago

What kind of function is 3y-2=x+7

Phoenix [80]

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Which of the following lines will have a negative slope? Select all that apply

mariarad [96]

I think it is 1, 4, and 5

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2 years ago

A student graphed f(x)=x and g(x)= f(x)-8 on the same coordinate grid. Which statement best describes how the graph of f and g a

luda_lava [24]

Answer:

The correct option is D

The graph of f is shifted 8 units down to create the graph of g.

Explanation:

Given f(x) = x

If the graph of this function is shifted 8 units down, we have

x - 8

This is the function f(x) - 8

Since g(x) = f(x) - 8, we conclude that the graph of f is shifted 8 units down to create the graph of g.

7 0

1 year ago

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago

Plz help with this question​

Plz help with this question