Please is can anybody help me i'll do any thing.

PLEASE HELP WILL GIVE BRAINLYEST AND 10 POINTS!!!

Alina [70]

Answer:

D

Step-by-step explanation:

6 0

3 years ago

Lynette buys candy at the grocery store. she gives half of the pieces to her son and 2 pieces to her daughter. if she has 4 piec

nevsk [136]

Let Lynette has originally candies in hand = x

She gave half to her son means x/2 given to son.

2 candies gave to daughter and left with 4.

x - x/2 - 2 = 4

adding 2 both sides

x - x/2 - 2 + 2 = 4 + 2

x - x/2 = 6

x /2 = 6

x = 12

Means she have 12 candies originally : Answer

Hope it will help :)

5 0

3 years ago

I need help with this 1 problem on my homework

Andru [333]

Step-by-step explanation:

10-5y=3x

1. add 10 to both sides

-5y=3x+10

2. divide everything by -5

y=-3/5x-2

3. your point would start at -2 on the y axis and you would go down 3 and over 5 to get to the next point

4 0

3 years ago

You deposit $200 in an account earning 3.5% simple interest how long will it take for the balance of the account to be $221

Rus_ich [418]

It takes 3 years for the balance of account to be $ 221

Solution:

From given information,

Principal = $ 200

Rate of interest = 3.5 %

Amount after "n" years = $ 221

To find: number of years

In simple Interest,

Simple interest = amount earned - principal

Simple interest = 221 - 200 = 21

Thus simple interest earned is $ 21

The formula for simple interest is given as:

$simple\ interest = \frac{ p \times n \times r}{100}$

Where,

p is the principal

n is the number of years

r is the rate of interest per annum

Substituting the given values in formula,

$21 = \frac{200 \times n \times 3.5}{100}\\\\21 = 2n \times 3.5\\\\n = \frac{21}{7}\\\\n = 3$

Thus it takes 3 years for the balance of account to be $ 221

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