Ask question

Ask question

All categories

3 years ago

5

If a line crosses though (5,-3) and (-1,6);what is the y-intercept?

1 answer:

harina [27]3 years ago

6 0

(5,-3) (-1,6)
y=mx+b

calculate m:
(-3-6)/(5-(-1)
=-9/6=-1.5

y=-1.5x+b

insert the (5, -3) point to calculate b:

-3=-1.5*5+b
-3=-7.5+b
4.5=b

b=the y intercept, so the line intercepts x=0 at y=4.5

You might be interested in

Jonathan purchased 4 shirts, 2 jackets, and 3 pants. How many different outfits using a

vagabundo [1.1K]

Answer:

24 way im pretty sure

Step-by-step explanation:

4 0

3 years ago

What is the gradient of the graph shown?

JulijaS [17]

Answer:

1

Step-by-step explanation:

5 0

2 years ago

Nathan plants equal sized squares of sod in his front yard. Each square has an area of 6 square feet.Nathan plants a total of 1,

Neko [114]

I THINK the answer is 6000 square feet because each one is 6 square feet and you have 1000 of them so 6 square feet times 1000 squares is 6000 square feet.

7 0

3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

3 years ago

Can someone please help!!!

Inga [223]

It is simple, each y=3x so, lets just say that you have 8y, that would be equal to 24x. In other words, just multiply the number of y's, by 3

3 0

3 years ago