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saveliy_v [14]
3 years ago
6

Can someone help please 2y+3=3(y+7)

Mathematics
2 answers:
vredina [299]3 years ago
8 0

Answer:

2y + 3 = 3(y + 7) \\ 2y + 3 = 3y + 21 \\ y \: temrs \: togather \\ 2y - 3y = 21 - 3 \\  - y = 18 \\ y =  - 18 \\ thank \: you

NikAS [45]3 years ago
5 0

Answer:

y = -18.

Step-by-step explanation:

hope that helps

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Please help! question is in the picture!
koban [17]

9514 1404 393

Answer:

  (5, 1) or maybe (h, k)

Step-by-step explanation:

The given form has no letters other than the variable. The vertex in this vertex-form equation can be read from the expression as (5, 1).

In the generic vertex form ...

  f(x) = a(x -h)² +k

the letters (h, k) are used to represent the vertex.

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You would need to consult your curriculum materials to see what letters your author uses for the vertex in vertex form.

5 0
3 years ago
Why is everyone posting inappropriate things??? Please stop!!! This is a website to help out people with their work!!!
Eddi Din [679]

Answer:

Agreed

Step-by-step explanation:

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3 years ago
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If y=2x- 3, find the value of y when x=2
Ira Lisetskai [31]

y = 2x - 3

y = 2 * 2 - 3

y = 4 - 3

y = 1

3 0
3 years ago
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
Help please I don’t understand
o-na [289]

Answer:

Look at the y-intercept and x-intercept

Step-by-step explanation:

Determine which one they're looking for by finding the intercept or slope.

5 0
3 years ago
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