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3 years ago

I need help with my homework

Mathematics

2 answers:

Advocard [28]3 years ago

6 0

Answer:

Step-by-step explanation:

Pythagorean Theorem: a²+b²=c², with a and b being the 2 shortest sides of the triangle, and c being the longest.

a² b² c²

12² + 35² = 1369

Once you have c², you need to find the square root of c² to get c.

The square root of 1369 is 37. c=37

Juli2301 [7.4K]3 years ago

3 0

The value of X=37

hope this helps :)

X= 12^2 + 35^2

= √1369

= 37

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Solve for xif mZRQS = 2x+4 and mZTQS = 6x+ 20.<br> 180<br> 19.5<br> -4<br> 90

Likurg_2 [28]

Answer:

x= -4

Step-by-step explanation:

2x+4=6x+20

so 2x-6x=20-4

divide both sides by -4

-4x/-4 = 16/4

=-4

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

How would u do this and can u explain

Amanda [17]

M=2-(-6)/(-1)-(-2)=8
Y=8x+b
-6=8*(-2)+b
-6=-16+b
B=16-6=10
Y=8x+10

I hope my answer is help to you

4 0

3 years ago

What is the answer to this math problem

shutvik [7]

Y + x = 5x + 3
12 - y = x + 2y

Isolate x in the first equation, then substitute its expression into the second equation to find y:

y + x = 5x + 3
x = 5x - y + 3
-4x = -y + 3
x = y - 3/4

---

12 - y = x + 2y
12 - y = (y - 3/4) + 2y
48 - 4y = y - 3 + 8y
48 - 4y = 9y - 3
-4y = 9y -51
-13y = -51
y = 51/13

Substitute this into the first equation to find x:

51/13 + x = 5x + 3
x - 5x = 3 - 51/13
-4x = -12/13
x = 3/13

Lastly, substitute back into both original equations to check work;

y + x = 5x + 3 --> 51/13 + 3/13 = 5(3/13) + 3 --> 54/13 = 54/13
12 - y = x + 2y --> 12 - 51/13 = 3/13 + 2(51/13) --> 105/13 = 105/13

Answer:
x = 3/13
y = 51/13

Hope this helps.

4 0

3 years ago

What set of transformations could be applied to rectangle ABCD to create A'B'C'D'?

sergeinik [125]

Answer:

3 or 4 i belive

Step-by-step explanation:

8 0

3 years ago