All categories

3 years ago

It can be equal to the set and it can contain all the elements that are present in the set

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Answer: yes

Step-by-step explanation:

You might be interested in

Parallelogram CDEF with vertices C(-4,-4), D(-2, 0), E(6, 1), and F(4, -3) in the line y = 2.

STALIN [3.7K]

Y=2 I’d the answer I think

7 0

3 years ago

The sixth grade art students are making a mosaic using tiles in shapes of right triangles. Each tiles has the leg measures of 5.

Drupady [299]

Given:

Each right triangular tiles has the leg measures of 5.2 cm and 6 cm.

There are 150 tiles in the mosaic.

To find:

The area of the mosaic.

Solution:

We know that, the area of a triangle is:

$A=\dfrac{1}{2}\times Base\times height$

So, the area of Each right triangular tile is:

$A=\dfrac{1}{2}\times 5.2\times 6$

$A=\dfrac{1}{2}\times 31.2$

$A=15.6$

There are 150 tiles in the mosaic. So, the area of the mosaic is:

$Area=15.6\times 150$

$Area=2340$

Therefore, the total area of the mosaic is 2340 cm ².

5 0

3 years ago

Z^4-4z^3+4z^2+48=0 how to find value of z?

Aliun [14]

There are no real solutions to solve Z. If there may be a typo or something, there are no solutions.

3 0

2 years ago

K^5/6<br> Verbal expression

fenix001 [56]

Answer:

The quotient of a number to the fifth power and six.

Step-by-step explanation:

Step 1: Convert k⁵ into words

k⁵ = a number to the fifth power

Step 2: Convert / into words

/ = quotient

Step 3: Convert 6 into words

6 = six

Step 4: Combine

The quotient of a number to the fifth power and six.

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