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2 years ago

Y" - 4y = ((x^2) - 3) sin 2x

Mathematics

1 answer:

Gekata [30.6K]2 years ago

8 0

Well i did this in my notes and i found this

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Solve 3x² + 4x = 2. (2 points)

uysha [10]

Answer:

Step-by-step explanation:

As it is a second order equation, it means that it has two possible answers and they are $x_1$ and $x_2$ .

The famous quadratic formula for solving any second order equation is the following:

$x_{1} = \frac{-b + \sqrt{b^2 - 4 ac}}{2a}\\x_{2}=\frac{-b - \sqrt{b^2 - 4 ac}}{2a}$

Where a is the coefficient of $x^2$ , b is the coefficient of x, and c is the free term. In other words,

$a = 3\\b=4\\c=-2$

as the equation should be in the following form:

$a x^2 + bx+c = 0$

Therefore the possible answer should be the following,

$\frac{-4 + \sqrt{4^2 - 4*3*(-2)}}{2*3}=\frac{-4 +\sqrt{16 + 24}}{6} =\frac{-4 + \sqrt{40}}{6}=\\ \frac{-4 + \sqrt{4*10}}{6} = \frac{-4 + 2\sqrt{10}}{6}$ $\frac{2*(-2 + \sqrt{10})}{2*3} = \frac{-2 + \sqrt{10}}{3}$

by dividing the numerator and denominator by 2, we can deduce the following,

6 0

1 year ago

you are baking muffins for your class. there are 17 total students in your class and you have baked 5 muffins. write and solve a

Tresset [83]

For your first answer 2=x+5/17 your second answer is 29. This is because "x" is the number that you need to bake which is 29. Therefore, 31/17=2

4 0

3 years ago

Read 2 more answers

PLZ HELP **URGENT** BRAINLIEST TO FIRST ANSWER **HELP PLZ**

kumpel [21]

Answer:

Step-by-step explanation:

EF - base1

HG - base2

J-midpoint of EH

K-midpoint of FG

We are solving it by using formula:

JK=(base1+base2):2

25=(3x+2+7x-2):2

25*2=10x

50=10x

x=50:10

x=5

Good luck!

5 0

3 years ago

Read 2 more answers

Sorry for the bother help plz

Zepler [3.9K]

(6a^2)^3

6^3(a^2)^3

216(a^2)^3

216a^6

Answer: 216a^6

3 0

3 years ago

Read 2 more answers

4 cos²x – 1=0<br><br> What is the solution?

andreev551 [17]

Answer:

$x=\frac{\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{5\pi }{3}+2\pi n, n \in \mathbb{Z}$

$x=\frac{2\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{4\pi }{3}+2\pi n, n \in \mathbb{Z}$

$x=\frac{\pi}{3}+\pi n, n \in \mathbb{Z}$

$x=\frac{2\pi }{3}+\pi n, n\in \mathbb{Z}$

Step-by-step explanation:

$4\text{cos}^2(x)-1=0\\4\text{cos}^2(x)=1\\$

$cos(x)=\pm\sqrt{\frac{1}{4} }$

$cos(x)=\pm\frac{1}{2}$

So, when cos(x) is equal to

$\frac{1}{2} \text{ and } -\frac{1}{2}$ ?

For

$cos(x)=\frac{1}{2}$

We are talking about x = 60º and x = 300º, Quadrant I and IV, respectively. In radians:

$x=\frac{\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{5\pi }{3}+2\pi n, n \in \mathbb{Z}$

$x=\frac{\pi}{3}+\pi n, n \in \mathbb{Z}$

For

$cos(x)=-\frac{1}{2}$

We are talking about x = 120º and x = 240º, Quadrant II and III, respectively. In radians:

$x=\frac{2\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{4\pi }{3}+2\pi n, n \in \mathbb{Z}$

$x=\frac{2\pi }{3}+\pi n, n\in \mathbb{Z}$

6 0

3 years ago