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2 years ago

14

Which expression illustrates the commutative property applied to the product?

1 answer:

kicyunya [14]2 years ago

3 0

Your answer should be c

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A fair coin is tossed 100,000 times. the number of heads is recorded. what is the probability that there are between 49,800 and

Natali5045456 [20]

Make it 49800/50000 and 50200/50000 and then compare.

4 0

2 years ago

What is the area of a circle with a diameter of 12.6 in.?

Zepler [3.9K]

Hey there! I'm happy to help!

To find the area of a circle, you square the radius and then multiply by pi (3.14 in our case).

The radius is half of the diameter.

12.6/2=6.3

We square this.

6.3²=39.69

We multiply by 3.14

39.69×3.14=124.6266

We round to the nearest hundredth, giving us an area of 124.63 in².

Now you can find the area of a circle! Have a wonderful day! :D

5 0

3 years ago

Solve 7x + 2 < 9x - 4

Alexus [3.1K]

Answer:

$\boxed{\bold{x>3}}$

Step By Step Explanation:

Subtract 2 From Both Sides

$\bold{7x+2-2$

Simplify

$\bold{7x$

Subtract 9x From Both Sides

$\bold{7x-9x$

Simplify

$\bold{-2x$

Multiply Both Sides By -1 (Reverse Inequality)

$\bold{\left(-2x\right)\left(-1\right)>\left(-6\right)\left(-1\right)}$

Simplify

$\bold{2x>6}$

Divide Both Sides By 2

$\bold{\frac{2x}{2}>\frac{6}{2}}$

Simplify

$\bold{x>3}$

- Mordancy

5 0

3 years ago

If you flip a fair coin 10 times and it lands heads up 5 times, this would result in 50% probability of the coin landing heads u

Shtirlitz [24]

This is an example of empirical probability, although it gives the same ratio as a theoretical probability, because we based the result on our experiment

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