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Pachacha [2.7K]

3 years ago

9

So I took a test from my teachers desk and took photos of the test questions. I am indeed cheating but I just want to know the a

nswers for the test before I take it tomorrow. Here is number 13.
It is Geometry PLEASE HELP!!!.

1 answer:

romanna [79]3 years ago

8 0

Answer:

THE CORRECT OPTION IS A) Δ ABC $\sim$ Δ BDC $\sim$ Δ ADB

Step-by-step explanation:

The first option is in correct order of triangle ,we can easily the their sides are proportionate . And having eqal angles.

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There are 10 girls and 8 boys in Mr Augellis class. If 5 more girls and 5 more boys join the class, will the ratio of girls to b

zimovet [89]

Yes they will stay the same

8 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

In order to unlock the mystery window, you must enter a 3 digit code. You are given these following: -The number can be divided

liraira [26]

343 or how to solve LCM OR GCF Since 1 divides into everything, then the greatest common factor in this case is just 1. When 1 is the GCF, the numbers are said to be "relatively" prime; that is, they are prime, relative to each other. Then the GCF is 1 and the LCM is 2 × 2 × 2 × 3 = 24.

I can’t seem to figure it out so sorry but example on how to find it ^

4 0

3 years ago

Since some people need the points, here’s a question where you’ll just need to answer it with anything and I give it 5 stars and

mezya [45]

Answer:

thank you very much

Step-by-step explanation:

4 0

3 years ago

5x+7y=31 2x+4y=16 Which of the following systems could be used to solve the given system of equations by the addition method? 10

Pani-rosa [81]

Equation 1: $5x+7y=31$
Equation 2: $2x+4y=16$

We can either use the 'elimination' method or the 'substitution' method to solve this simultaneous equation.

In some cases one method is easier to use than the other. For this one, it would be easier to use the elimination method

We need to eliminate either the 'x' or the 'y' to begin with. Say we want to eliminate the 'x' terms, then the next step is to make the constant the same

Equation 1: the constant of 'x' is 5
Equation 2: the constant of 'x' is 2

To make the two constants the same, think of common multiple. The lowest common multiple for 5 and 2 is 10

Equation 1: Multiply all terms by 2 to achieve 10x
Equation 2: Multiply all terms by 5 to achieve 10x

Equation 1: $10x+14y=62$
Equation 2: $10x+20y=80$

Once the constant of 'x' is the same, then we can start the elimination process. Since our aim is to eliminate, one of the 'x' need to be in negative form. We can either multiply Equation 1 or Equation 2 by (-1)

Equation 1: -10x - 14y = -62
Equation 2: 10x + 20y = 80

Hence the correct answer is: Third option

3 0

4 years ago