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3 years ago

10

So I need someone to asnwer this ASAp since my he is due in 20 minutes. Please help!

1 answer:

dem82 [27]3 years ago

5 0

A)6 ^^^^^^^^^^^^because

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Find the value of f(12) if f(x)=-2x+4

Contact [7]

Answer:

f(12) = - 20

Step-by-step explanation:

To evaluate f(12), substitute x = 12 into f(x), that is

f(12) = - 2(12) + 4 = - 24 + 4 = - 20

5 0

2 years ago

Solve for X. Please!!

pychu [463]

Answer:

x = 37.5

Step-by-step explanation:

By Basic Proportionality Theorem:

$\frac{x}{20} = \frac{46 - 16}{16} \\ \\ \frac{x}{20} = \frac{30}{16} \\ \\ x = \frac{20 \times 30}{16} \\ \\ x = \frac{600}{16} \\ \\ x = 37.5$

3 0

2 years ago

Factor the following expression. Simplify your answer.<br> 3s(s - 1)^1/3 + 2(s - 1)^4/3

denis23 [38]

Answer:

Step-by-step explanation:

$3s\sqrt[3]{s-1} + 2 \sqrt[4/3]{s-1} =\\3s\sqrt[3]{s-1} + 2 \sqrt[1/3]{(s-1)^4} =\\3s\sqrt[3]{s-1} + 2 (s-1)\sqrt[1/3]{s-1} =\\\sqrt[3]{s-1}*(3s + 2 (s-1)) =\\\sqrt[3]{s-1}*(3s + 2s-2)) =\\\sqrt[3]{s-1}*(5s -2) \\$

6 0

2 years ago

Read 2 more answers

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

Solve the exponential equation for x. 256=(1/4)^3r+2

Vitek1552 [10]

Answer:

4^4= 4^-1(3r+2)

4^4= 4^(-3r-2)

-3r-2 = 4

-3r= 6

r= -2

-

Step-by-step explanation:

3 0

3 years ago