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yulyashka [42]
3 years ago
8

What is the sum of the polynomials?

Mathematics
1 answer:
kiruha [24]3 years ago
6 0
I don’t see the picture
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13x+5y=90 what is x? what is y?
Leona [35]

Answer:

X=-5y/13+90/13, Y=-13x/5+18

Step-by-step explanation:

You need to solve the equation for X and Y.

Solving for X:

13x+5y=90

Subtract 5y: 13x=-5y+90

Divide by 13: x=-5y/13+90/13

--

For Y:

13x+5y=90

Subtract 13x: 5y=-13x+90

Divide by 5: y=-13x/5+90/5

(Simplifies to -13x/5+18)

3 0
3 years ago
Extra points <br><br> not in words
Mashcka [7]
Step^1: 6 - 1 - 2(n) + 5 \\ Step^2: 2(n) = 0 \\ Step ^3: \frac{2n}{0} = 0 ; \\ Step^4 \frac{0}{0} = 0 \\ Step^5:Answer : n = 0 \\ Step^6: Number Of Solutions: 1Solution
good\\luck\\on\\your \\ assignment
6 0
3 years ago
please help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111 -9 2/7 - (-10 3/7) I NEED IT AS A FRACTION
mario62 [17]
8/7 exact form
or
 1 1/7 mixed number form
7 0
3 years ago
This is stupid but if I have 90x how do I get 90?
Marrrta [24]

Answer:

you should subtract x.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
In this problem, y = c1ex + c2eâx is a two-parameter family of solutions of the second-order DE y'' â y = 0. Find a solution of
SCORPION-xisa [38]

Answer:

y(t) = 2e^t -e^{-t}

Step-by-step explanation:

Assuming this complete problem: "In this problem,

y = c1ex + c2e−x

is a two-parameter family of solutions of the second-order DE

y'' − y = 0.

Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

y(0) = 1,    y'(0)= 3"

Solution to the problem

For this case we have a homogenous, linear differential equation with order 2, and with the general form:

ay'' +by' +cy=0

Where a =1, b=0, c=-1

And we can rewrite the differential equation in terms y = e^{rt} like this:

[e^{rt}]'' -e^{rt}=0

And applying the second derivate we got:

r^2 e^{rt} -e^{rt}=0

We can take common factor e^{rt} and we got:

e^{rt} (r^2-1) =0

And for this case the two only possibel solutions are r=1, r=-1

And the general solution for this case is given by:

y = c_1 e^{r_1 t} + c_2 e^{r_2 t}

Replacing the roots that we found we got:

y = c_1 e^{t} +c_2 e^{-t}

Now we can find the derivates for this last espression

y' = c_1 e^t -c_2 e^{-t}

y'' = c_1 e^t +c_2 e^{-t}

From the initial conditions we have this:

y(0)=1 =c_1 e^{0} +c_2 e^{-0}= c_1 +c_2   (1)

y'(0) =3= c_1 e^{0} -c_2e^{-0}= c_1 -c_2   (2)

If we add equations (1) and (2) we got:

4 = 2c_1 , c_1 = 2

And solving for c_2 we got:

c_2=3-c_1= 3-2 = 1

So then our general solution is given by:

y(t) = 2e^t -e^{-t}

8 0
3 years ago
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