All categories

3 years ago

Please help it’s due now

Mathematics

1 answer:

insens350 [35]3 years ago

3 0

Answer:

y=6x+9

Step-by-step explanation:

y=mx+b

add 6x on both sides

You might be interested in

Evaluate the expression: v ? w Given the vectors: r = <8, 1, -6>; v = <6, 7, -3>; w = <-7, 5, 2> . If the answ

Snezhnost [94]

Answer:

-13

Step-by-step explanation:

Given are two vectors v and w.

v=(6,7,-3) and

w=(-7,5,2)

We are to find the dot product of v.w

We have

Dot product is obtained by multiplying corresponding pairs and adding them

Here we have

v.w=6(-7)+7(5)-3(2)

=-13

7 0

3 years ago

A bag of pecans costs $3.84. If

Stels [109]

Answer:

10.4 cents

Step-by-step explanation:

40 pecans/$3.84 gives you pecans per dollar

6 0

4 years ago

Can someone help me with this problem

stepan [7]

The answer is D. 3 and 17.
3 + 17 = 20. They have a sum of 20.
17 - 3 = 14. They also have a difference of 14.
Hope I helped!

7 0

3 years ago

5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

5 0

4 years ago

14. Is x + 2 a factor of the polynomial f(x)=2x-3x²-4x+1?

defon

Thanks for the free 10x+ points .

8 0

3 years ago