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2 years ago

What numbers should i put in the drop down menus? the numbers i can choose are 0-9.

Mathematics

1 answer:

ElenaW [278]2 years ago

6 0

Answer:

1x+4 should be the answer

Step-by-step explanation:

6x-3x-2x=1x

4=4

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Find the product. (a2 + 2a + 1)(a + 1) a. A2 + 3a + 2 b. A3 + 3a2 + 3a + 1 c. A3 + 2a 2 + a + 1 d. A3 + 3a2 + a + 1

olchik [2.2K]

(a + 1)(a² + 2a + 1)

= a(a² + 2a + 1) + 1(a² + 2a + 1)

= a³ + 2a² + a + a² + 2a + 1

= a³ + 3a² + 3a + 1

Answer: b

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2 years ago

Tim needs 23 3/4 m of wood trim to remodel a room. Wood trim is sold in lengths of 1 1/4 m. What equation can be used to find th

Bumek [7]

I hope this helps you

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2 years ago

5/ 10 of the students in Sam’s class are girls. Of the girls, 1 /5of them wear glasses. What fraction of all the students wear g

Vesnalui [34]

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2 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,

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

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2 years ago

7b+1=-53 taking square roots

Aliun [14]

7.28 is the answer sorry if i’m wrong

6 0

2 years ago