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2 years ago

Can someone answer this

Mathematics

1 answer:

attashe74 [19]2 years ago

6 0

Answer:

Step-by-step explanation:

axis: x= 2

vertex: (2,3)

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Determine the value of x in the diagram.<br> x<br> 2x<br> 2x

ryzh [129]

Answer:

x = 60

Step-by-step explanation:

The sum of exterior angles in a rectangle is equal to 360

so to find the value of x we need to use all given values and write an equation:

x + 2x + x + 2x = 360 add like terms

6x = 360 divide both sides by 6

x = 60

5 0

1 year ago

If there are 16 boys and 32 girls in a room, fill out all of the possible ratios of boys to girls that could be made. They alrea

Rzqust [24]

All of possible ratios of boys to girls that could be made is 8:16; 4:8; 2:4; 1:2
Hope it helps, Brainliest if it <3

8 0

2 years ago

¿Alguien sabe como realizarlo?

sashaice [31]

Hola! Intente resolverlo pero solo hice la primera por mi tiempo ahorita mismo y era:
f(0)=1
Toda la ecuación

7 0

2 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

2 years ago

Arithmetic of Functions problem.

lidiya [134]

Answer:

(f o g)(4) = 45

Step-by-step explanation:

f(x)=4x+1

g(x)=x²-5

(f o g)(4)=?

(f o g)(4) = f(g(4))

Calculating g(4):

x=4→g(4)=4²-5

g(4)=16-5

g(4)=11

Replacing g(4)=11

(f o g)(4) = f(g(4))

(f o g)(4) = f(11)

Calculating f(11)

x=11→f(11)=4(11)+1

f(11)=44+1

f(11)=45

Replacing f(11)=45:

(f o g)(4) = f(11)

(f o g)(4) = 45

4 0

3 years ago

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