PLEASE, I REALLY NEED HELP!!!

Mathematics

1 answer:

Masja [62]4 years ago

5 0

50 - 55 million year ago. And I think for the first one reptiles were around.

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Decrease $7500 by 7%<br>please help i need step by step explanation

soldier1979 [14.2K]

(100% - 7%) × 7,500

= 93% × 7,500

= 93 ÷ 100 × 7,500

= 93 × 7,500 ÷ 100

= 697,500 ÷ 100

= 6975
hope this helped

6 0

3 years ago

Read 2 more answers

Write 9x² - 36x = 4y² + 24y + 36 in standard form

Kisachek [45]

Answer:

9·x² - 36·x = 4·y² + 24·y + 36 in standard form is;

(x - 2)²/2² - (y + 3)²/3² = 1

Step-by-step explanation:

The standard form of a hyperbola is given as follows;

(x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/b² - (x - h)²/a² = 1

The given equation is presented as follows;

9·x² - 36·x = 4·y² + 24·y + 36

By completing the square, we get;

(3·x - 6)·(3·x - 6) - 36 = (2·y + 6)·(2·y + 6)

(3·x - 6)² - 36 = (2·y + 6)²

(3·x - 6)² - (2·y + 6)² = 36

(3·x - 6)²/36 - (2·y + 6)²/36 = 36/36 = 1

(3·x - 6)²/6² - (2·y + 6)²/6² = 1

3²·(x - 2)²/6² - 2²·(y + 3)²/6² = 1

(x - 2)²/2² - (y + 3)²/3² = 1

The equation of the hyperbola is (x - 2)²/2² - (y + 3)²/3² = 1.

5 0

3 years ago

Find a particular solution to the nonhomogeneous differential equation y'' + 4 y = cos(2x) + sin(2x).

alukav5142 [94]

The characteristic solution follows from solving the characteristic equation,

$r^2+4=0\implies r=\pm2i$

so that

$y_c=C_1\cos2x+C_2\sin2x$

A guess for the particular solution may be $a\cos2x+b\sin2x$ , but this is already contained within the characteristic solution. We require a set of linearly independent solutions, so we can look to

$y_p=ax\cos2x+bx\sin2x$

which has second derivative

${y_p}''=(-4ax+4b)\cos2x+(-4bx-4a)\sin2x$

Substituting into the ODE, you have

$y''+4y=\cos2x+\sin2x$
$\implies4b\cos2x-4a\sin2x=\cos2x+\sin2x$
$\implies\begin{cases}4b=1\\-4a=1\end{cases}\implies a=-\dfrac14,b=\dfrac14$

Therefore the particular solution is

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

Note that you could have made a more precise guess of

$y_p=(a_1x+a_0)\cos2x+(b_1x+b_0)\sin2x$

but, of course, any solution of the form $a_0\cos2x+b_0\sin2x$ is already accounted for within $y_c$ .