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1 year ago

There are 60 students in Kinsley's grade. 3 of the students are enrolled in health. What

1 answer:

4 0

5% of the kids are in health. I got this from simplifying 3/60 which is 1/20

After that, i set up this equation: 1/20 times x/100

I would then divide 20 and 100 to get 5 and that’s basically what I did

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Complete this statement: 20 a x 2 + 25 a x + 15 a = 5 a 20 a x 2 + 25 a x + 15 a = 5 a

liubo4ka [24]

Answer:

Step-by-step explanation:

(((22•5ax2) + 25ax) + 15a) - 5 = 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

20ax2 + 25ax + 15a - 5 =

5 • (4ax2 + 5ax + 3a - 1)

Equation at the end of step 3 :

5 • (4ax2 + 5ax + 3a - 1) = 0

Step 4 :

Equations which are never true :

4.1 Solve : 5 = 0

4 0

3 years ago

Help Please

yaroslaw [1]

Answer:

Its (10,∞)

Step-by-step explanation:

Trust me I just took the exam and got it correct

5 0

3 years ago

Your friend comes from a large family. This list shows the ages of all of the children in her family. What is the median age of

Lyrx [107]

Answer:

It is 11 and 4/7 I am pretty sure hope I helped!

5 0

2 years ago

2y-x=6 and y=2x+7 graphed equals how many solutions

zhannawk [14.2K]

Answer:

One solution

<em>BRAINLIEST, PLEASE!</em>

Step-by-step explanation:

2y - x = 6

2y = x + 6

y = x/2 + 3

y = 2x + 7

After graphing, they have one solution at (-2.667, 1.667).

8 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:

$y_{p}(t) = C$

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

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

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:

$y_{p}(t) = C$

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

$(C)' + 4C = 8$

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