Ask question

Ask question

All categories

2 years ago

15

Which values of a, b, and c represent the answer in simplest form?

1 answer:

max2010maxim [7]2 years ago

5 0

Answer:

Which values of a, b, and c represent the answer in simplest form?

StartFraction 9 Over 11 EndFraction divided by StartFraction 5 Over 11 EndFraction = a StartFraction b Over c EndFraction

a = 1, b = 99, c = 55

a = 1, b = 55, c = 99

a = 1, b = 4, c = 5

a = 1, b = 5, c = 4

Step-by-step explanation:

You might be interested in

You have one type of nut that sells for $4.50/lb and another type of nut that sells for $8.00/lb. You would like to have 28 lbs

-Dominant- [34]

Answer:

I will need 26.4 pounds of the expensive nuts and 1.6 pounds of the cheap nuts.

Step-by-step explanation:

Since I have one type of nut that sells for $ 4.50 / lb and another type of nut that sells for $ 8.00 / lb, and I would like to have 28 lbs of a nut mixture that sells for $ 7.80 / lb, to determine how much of each nut will I need to obtain the desired mixture, the following calculation must be performed:

8 x 0.95 + 4.5 x 0.05 = 7.825

8 x 0.94 + 4.5 x 0.06 = 7.79

0.94 x 28 = 26.32

26.4 x 8 + 1.6 x 4.5 = 218.4

218.4 / 28 = 7.8

Thus, I will need 26.4 pounds of the expensive nuts and 1.6 pounds of the cheap nuts.

7 0

2 years ago

notka56 [123]

Answer:

yes razi is correct pateran 3

Step-by-step explanation:

8 0

3 years ago

21:28 = :24<br><br>plz help....

gavmur [86]

Answer:

x=18

Step-by-step explanation:

21:28=x:24

1. rewrite: 21/28 = x/24

2. simplify: 3/4=x/24

3. multiply 24 to both sides: x=18

4 0

2 years ago

If a dozen eggs cost $1.35, what is the unit cost?

Ulleksa [173]

Total eggs=12
dozen eggs cost =1.35 dollars
one egg =12/1.35
=0.11 dollars

4 0

2 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