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3 years ago

14

Ray says that the ordered pair for point G is(1,2).Is Ray correct?

1 answer:

liubo4ka [24]3 years ago

4 0

Answer:

1 unit up from the origin, so the ordered pair is (2, 1). (4, 2); Possible explanation: Point H is 4 units to the right of the origin, so the x -coordinate is

NO hes not correct

Step-by-step explanation:

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Is 26,341 divisible by 9

Paul [167]

No, it would leave a decimal

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2 years ago

Help! no links please!

Simora [160]

Answer: D

Step-by-step explanation:

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2 years ago

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will the opposite of a number always,sometimes,or never be greater than the number itself??????? can someone do an example and a

weeeeeb [17]

When the number is positive, its opposite is less than the number.

9 ==> -9 . . . . . . less
72 ==> -72 . . . . less
324 ==> -324 . . less

When the number is negative, its opposite is greater than the number.

-8 ==> 8 . . . . . . greater
-3/4 ==> 3/4 . . . greater
-429 ==> 429 . . greater .

We've seen that the opposite can be less, and it can also be greater.
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2 years ago

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Help me solve this questions with steps please

Flauer [41]

Answer:

B

Step-by-step explanation:

Since the triangle is right use the tangent ratio to find x

tan22° = $\frac{opposite}{adjacent}$ = $\frac{203}{x}$

Multiply both sides by x

x × tan22° = 203 ( divide both sides by tan22° )

x = $\frac{203}{tan22}$ ≈ 502.5 m → B

7 0

3 years ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago