How do I find a Slope for ( 2,2)and (-3,1)

Molly is excited to have fresh basil at home. She buys a basil plant that is 90 millimeters tall and puts it on her kitchen wind

satela [25.4K]

Answer:

Molly's basil plant is now 17 centimeters

Step-by-step explanation:

90 millimeters=9 centimeters

20cm-12cm=8cm

9cm+8cm=17cm

hope this helped :)

5 0

1 year ago

Solve the following proportion problem for x: 14/3x=20/x-5

Butoxors [25]

14 20

---- = ----- cross multiply

3x x-5

14(x-5) = 3x (20)

14x - 70 = 60x

-14x -14x

-70 = 36x

/36 /36

-70/36 = x

or

-35/18 = x

5 0

3 years ago

Read 2 more answers

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

If f(z)=m^z - n^z, and f(1)=2, f(2)=8. Find f(3).

Marina CMI [18]

If f(z)=m^z - n^z <span>and f(1)=2, f(2)=8
then
m - n = 2 so m = n + 2
m^2 - n^2 = 8

substitute </span>m = n + 2 into m^2 - n^2 = 8

so
m^2 - n^2 = 8
(n + 2)^2 - n^2 = 8
n^2 + 4n + 4 - n^2 = 8
4n = 4
n = 1

m = n + 2 = 1 + 2 = 3

so f(z) = 3^z - 1^z

if z = 1 the f(1) = 3^1 - 1^1 = 2
if z = 2 the f(2) = 3^2 - 1^2 = 9 - 1 = 8
if z = 3 the f(3) = 3^3 - 1^2 = 27 - 1 = 26

answer
f(3) = 26

3 0

3 years ago

Select the two binomials that are factors of this trinomial.

Free_Kalibri [48]

Answer:

(x-5)(x+4)

Step-by-step explanation:

hope it helps

3 0

3 years ago