All categories

3 years ago

Please help and add explanation

Mathematics

1 answer:

Degger [83]3 years ago

3 0

Answer:

y=-3x+10

Step-by-step explanation:

Since you know that the equation of the line is perpendicular to 1/3, you have to find the inverse of 1/3. You swap the denominator and numerator and make the number negative. In this case, it will be -3. Now that you know the slope is -3, you can use the equation y-y1=m(x-x1) where m is the slope. We know that the point (2,4) is in the graph so we can plug it into the equation. y-4=m(x-2). We know that the slope is -3 so we can plug it into the equation. y-4=-3(x-2). Simplify this to get y-4=-3x+6. Simplify this to get y=-3x+10. This means that the answer is y=-3x+10.

If this has helped you please mark as brainliest

You might be interested in

What is 8.51 x 104 in standard form?

notsponge [240]

I think the answer would be 85.1

8 0

3 years ago

Read 2 more answers

Someone help me please!!!!

Ghella [55]

I am pretty sure number 3 is the right answer

8 0

3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

Help once again. this is 6th grade work so uhh help

dolphi86 [110]

The answer is that s = 55/42.

Here's how to solve it

3 0

2 years ago

Read 2 more answers

Rewrite in polar form: x^2+y^2-2y=7

Alchen [17]

X²+y²-2y=7
using the formula that links Cartesian to Polar coordinates
x=rcosθ and y=r sin θ
substituting into our expression we get:
(r cos θ)²+(r sin θ)²-2rsinθ=7
expanding the brackets we obtain:
r²cos²θ+r²sin²θ=7+2rsinθ
r²(cos²θ+sin²θ)=7+2rsinθ
using trigonometric identity:
cos²θ+sin²θ=1
thus
r²=2rsinθ+7
Answer: r²=2rsinθ+7

8 0

3 years ago

Read 2 more answers