All categories

3 years ago

12. MULTIPLE CHOICE Brandon is watching a movie

Mathematics

1 answer:

frozen [14]3 years ago

5 0

Answer:

this problem has an answer.

Step-by-step explanation:

Boom easy points

IDRK I'd say D

You might be interested in

I need to solve this 1/4(-8y)-3x+9y-x

Alja [10]

Hope it helps to solve it

5 0

3 years ago

What's the answer to 1&2 plzz

Dmitrij [34]

1. 2880
2. 67%
Its simple math just put in effort

4 0

3 years ago

6 poodles to 18 beagles

JulijaS [17]

Answer:WHY YOU CHEATING LOL

Step-by-step explanation:

3 0

3 years ago

I DONT KNOW THESE! HELP!

kkurt [141]

Answer:

20. x = 5, y = -2

21. x = 11, y = 12

22. x = 22, y = 11

23. x = 11, y = 10

Step-by-step explanation:

20. Opposite sides in the parallelogram are congruent, so

$2y+18=3x-1\\ \\6x-3=17-5y$

Solve this system of two equations:

$\left\{\begin{array}{l}2y-3x=-19\\ \\6x+5y=20\end{array}\right.$

Multiply the first equation by 2 and add two equations:

$2(2y-3x)+6x+5y=2\cdot (-19)+20\\ \\4y-6x+6x+5y=-38+20\\ \\9y=-18\\ \\y=-2$

Substitute it into the first equation:

$2\cdot (-2)-3x=-19\\ \\-3x=-19+4\\ \\-3x=-15\\ \\x=5$

21. Opposite angles in the parallelogram are congruent, so

$11x+5=10y+6$

Consecutive angles are supplementary, so

$6x-y+11x+5=180^{\circ}$

Solve this system of two equations:

$\left\{\begin{array}{l}11x-10y=1\\ \\17x-y=175\end{array}\right.$

From the second equation

$y=17x-175$

Substitute it into the first equation:

$11x-10(17x-175)=1\\ \\11x-170x+1750=1\\ \\-159x=-1749\\ \\x=11\\ \\y=17\cdot 11-175=187-175=12$

22. Opposite angles in the parallelogram are congruent, so

$2x-5=3y-12$

Consecutive angles are supplementary, so

$2x-5+7y+x=180^{\circ}$

Solve this system of two equations:

$\left\{\begin{array}{l}2x-3y=-7\\ \\3x+7y=185\end{array}\right.$

From the first equation

$x=-3.5+1.5y$

Substitute it into the second equation:

$3(-3.5+1.5y)+7y=185\\ \\-10.5+4.5y+7y=185\\ \\-105+45y+70y=1,850\\ \\115y=1,850+105\\ \\115y=1,955\\ \\y=17\\ \\x=-3.5+1.5\cdot 17=22$

23. Opposite sides in the parallelogram are congruent, so

$2x+9=4y-9\\ \\3x-5=2y+8$

Solve this system of two equations:

$\left\{\begin{array}{l}2x-4y=-18\\ \\3x-2y=13\end{array}\right.$

Multiply the second equation by 2 and subtract it from the first equation:

$2x-4y-2(3x-2y)=-18-2\cdot 13\\ \\2x-4y-6x+4y=-18-26\\ \\-4x=-44\\ \\x=11$

Substitute it into the first equation:

$2\cdot 11-4y=-18\\ \\-4y=-18-22\\ \\-4y=-40\\ \\y=10$

3 0

3 years ago

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