All categories

3 years ago

???? HELPPPPPPPPPPPPPPPPPP

Mathematics

1 answer:

guapka [62]3 years ago

8 0

Plot points through plugging in a "x" to generate a "y".

y = 4x - 2

when x = 0, y = -2 (because 4 x 0 = 0, 0 - 2 = -2)
when x = 1, y = 2 (because 4 x 1 = 4, 4 - 2 = 2)
when x = 2, y = 6 (because 4 x 2 = 8, 8 - 2 = 6)
when x = 3, y = 10 (because 4 x 3 = 12, 12 - 2 = 10)

so and and so forth

after you find the points, connect the dots, and you will get your graph

hope this helps

You might be interested in

Explain how you can check your solution<br>to the equation. Then check the solution.

Hoochie [10]

Answer:

To check if a given value is a solution to an equation:

Evaluate the left-hand side expression at the given value to get a number.

Evaluate the right-hand side expression at the given value to get a number.

See if the numbers match.

4 0

3 years ago

I bought a carpet for my bedroom. The length of the carpet was 6 ft. The width was 12 ft. How many square feet is my carpet?

8090 [49]

72 square feet

EXPLANATION: when finding area of a rectangle, multiple Length x Width and 6x12=72

5 0

3 years ago

A school system is reducing the amount of dumpster loads of trash removed each week. In week 5, there were 60 dumpster loads of

Lana71 [14]

Answer:

f(x) = -4x + 80

Step-by-step explanation:

You are given two points,

(5, 60) and (10, 40)

in order to get the equation, let's use the form slope-intercept form

m = (y1 - y2) / (x1 - x2)

m = (60 - 40) / (5 - 10)

m = 20/-5

m = -4

Get the x intercept

y = mx + b

60 = (-4)(5) + b

b = 60+20

b = 80

so the equation is

y = -4x + 80

f(x) = -4x + 80

5 0

3 years ago

Two roads cross at an intersection. If one of the angles created is 50 degrees what’s are the three angles

Natasha2012 [34]

50 degrees is one while the other two are 130 degrees

4 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

4 years ago