Your school day begins at 8:50 a.m. and ends at 3:10 p.m. How long are you in school?

Write the equation of a line that passes through points (0,4) and (-2,-3) in slope intercept form

Otrada [13]

y=mx+b is the equation of a line;

m=slope , b= y-intercept

You can find the slope with this following equation: (y(2)-y(1))/(x(2)-x(1))

In this case the points are (0,4) and (-2,-3). The first set being (0,4) and the second (-2,-3). This means (0,4) can be expressed as (x(1),y(1)) and (-2,-3) expressed as (x(2),y(2)). Plugging these numbers into the slope equation gives us: (-3-4)/(-2-0) = -7/-2 = 7/2.

m= 7/2 ; so we have : y= (7/2)x+b

We are give a set of points which it passes through, we can simply plug them in:

4 = (7/2)(0)+b (0 is the x and 4 is the y)

We get 4 = 0 +b .... 4=b

our final equation is : y=(7/2)x+4

4 0

3 years ago

What is <br> 8v+w+7−8v+2w

Bingel [31]

Answer:

3w+7

Step-by-step explanation:

Let's simplify step-by-step.

8v+w+7−8v+2w

=8v+w+7+−8v+2w

Combine Like Terms:

=8v+w+7+−8v+2w

=(8v+−8v)+(w+2w)+(7)

=3w+7

3 0

2 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago

The irrational number square root 67 is less than [blank]

Solnce55 [7]

Answer:

The square root of 67 is less than the whole number of 9

Step-by-step explanation:

8 squared is 64

9 squared is 81

67 is in between them

4 0

2 years ago

Re write the expression 8^6 x 8^3/ 8^5 as an exponential expression with a single base

Arada [10]

Answer:

8^4

Step-by-step explanation:

Here we have:

8^6 · 8³

-------------

8^5

That's multiplication in the numerator. The appropriate rule of exponents states that:

8^a·8^b = 8^(a+b), so the numerator is equivalent to 8^(6 + 3) = 8^9.

Now we have

8^9

------

8^5

and the appropriate rule for division here is

8^a / 8^b = 8^(a - b)

So our:

8^9

--------- 8^(9-5) = 8^4

8^5

4 0

2 years ago