Equivalent expression to −5.55−8.55c+4.35c

1 answer:

5 0

-5.55-8.55c+4.35c 2nd and 3rd term is like term we can connect. -digit have to larger number than +digit so the sign will be negative
-5.55-4.2c there is no like term to connect
Hence this is the answer

You might be interested in

How many solutions does the system have?

sertanlavr [38]

If a solution(s) exists y=y so we can say:

x^2-3x=-2x+2 add 2x to both sides

x^2-x=2 subtract 2 from both sides

x^2-x-2=0 factor

(x-2)(x+1)=0

So x=-1 and 2, using y=-2x+2 we find:

y(-1)=4 and y(2)=-2

So the two solutions occur at the points:

(-1,4) and (2,-2)

4 0

4 years ago

Read 2 more answers

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

How do you write a word problem using 5621÷23

zepelin [54]

Megan has $5,621 to give to charity. She wanrs to give the money to 23 different charities. Each charity will get the same amount. How much money will each charity get? How much will be left over? Answer: 22r9

4 0

3 years ago

20 yd<br> .<br> 12 yd<br> square yards<br> Submit

kicyunya [14]

Answer:

140yd2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

How many solutions does this nonlinear system of equations have?

bulgar [2K]