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2 years ago

What angle corresponds with angle A?

Mathematics

1 answer:

jolli1 [7]2 years ago

8 0

The angle that corresponds with angle A will be P on the other triangle. Both angle A and P are 68 degrees.

And if you look at the sides of the triangle there is a similarity, proving the sides to be similar to each other.

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What must you invest today to receive an annuity of $4,500 for 10 years compounded at 8% quarterly when all withdrawals will be

OlgaM077 [116]

Answer:

Kent is making a scale model of his favorite train. The actual train is 12 feet long and 4 feet wide. Kent wants his model to be 6 inches in length. Which would if he uses the same ratio?

Step-by-step explanation:

Kent is making a scale model of his favorite train. The actual train is 12 feet long and 4 feet wide. Kent wants his model to be 6 inches in length. Which would be the width if he uses the same ratio?

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2 years ago

Factor over the complex numbers, if possible: x² + 2xyi - y2<br><br> What is the answer?

yanalaym [24]

9514 1404 393

Answer:

(x +yi)²

Step-by-step explanation:

Comparing the given expression to the square of a binomial, we can determine the given expression to be a square.

(a +b)² = a² +2ab +b²

Here, we have a=x and b=yi. Then the factorization is ...

x² +2xyi +(yi)² = (x +yi)²

8 0

2 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

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