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3 years ago

What are x- intercepts and y- intercepts

Mathematics

1 answer:

Trava [24]3 years ago

6 0

Answer: x and y intercepts are where the line crosses over the x and y axis.

Step-by-step explanation:

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How do you simplify:

andreev551 [17]

So 10.
first parenthasees ()
so 10(2.1q-2)=21q-10
then you do the rest

21q-20+7.5+18
add like terms
28.5q-2

second
(-6y-2)+5(3+2.5y)
distribute the 5
15+12.5y
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-6y-2+15+12.5y
6.5y+13

3rd
1/2(6x+10)+1/3(6+9x)
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3x+5+2+3x
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6x+7

4th
(3x+8)-6(2.5x-3)
distrubute
-15x+18
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3x+8-15x+18
-12x+26

5th
(7-3t)-5(-1.6t+5)
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+8t-25
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7-3t+8t-25
5t-18

6th
1/2(12w+8)-1/5(10w-5)
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4w+5

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

What is 1/4 of a 100 yard by 60 yard field

Tema [17]

Ok, im not going to answer it, but I'm gonna tell you what you need to do to get the answer, first, multiply 100 with 60, then take a quarter out, this is a very simple way to do it, and most people just overthink things.

8 0

4 years ago

Julio makes a scale drawing of the kitchen floor to determine how

tatuchka [14]

The scale drawing of Julio's kitchen would change the kitchen size

The actual area is 243 square feet

<h3>How to determine the actual area of Julio's kitchen?</h3>

The parameters are given as:

Length = 4.5 inches
Width = 3.375 inches

Start by converting the dimensions to feet using the scale ratio

Actual length = 4.5 inches/ (1 inch/ 4 feet) = 18 feet
Actual width = 3.375 inches/ (1 inch/ 4 feet) = 13.5 feet

The area is then calculated as:

Actual Area = Actual length * Actual Width

This gives

Actual Area =18 feet * 13.5 feet

Evaluate the product

Actual Area = 243 square feet

Hence, the actual area is 243 square feet