All categories

2 years ago

ILL GIVE BRAINLIEST TO THE CORRECT ANSER BUT ANSWER ASAP

Mathematics

2 answers:

Nezavi [6.7K]2 years ago

7 0

The answer is 35, 0,1,4,9 etc are all square numbers

alina1380 [7]2 years ago

6 0

The answer is 35

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers.

You might be interested in

How do you find x and y on the hypotenuse and long side of a roght triangle

DanielleElmas [232]

By pythagorean thoerm u can find
Hy^2=length^2+base^2

8 0

3 years ago

Chips suppose a computer chip manufacturer rejects 2% of the chips produced because they fail presale testing.

Artyom0805 [142]

(a) P( fifth one is bad) = P( first 4 are OK) * P(5th is bad)

= (0.98)^4 * 0.02 = 0.0184 or 1.84%

(b) this will be (0.98)^10 = 81.70%

4 0

3 years ago

Find the complete factored form of the polynomial: 6xy - 8x2

Ray Of Light [21]

Answer:

2x(3y-4x)

Step-by-step explanation:

6xy - 8x^2

Factor out the greatest common factor of 2x

2x*3y -2x*4x

2x(3y-4x)

7 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

Solve: 16x = 8 A. 1/2 B. 2 C. 4 D. 12

Rudik [331]

X=1/2
Hope this helped! :3

3 0

3 years ago

Read 2 more answers