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Oduvanchick [21]
3 years ago
9

Need Help ....!!!

Mathematics
1 answer:
postnew [5]3 years ago
4 0

\\ \sf\longmapsto \dfrac{(0.0225)^{\frac{-3}{2}}\times (0.0001)^{\frac{3}{4}}}{(0.0125)^{\frac{1}{3}}}

\\ \sf\longmapsto \dfrac{((0.15)^2)^{\frac{-3}{2}}\times ((0.1)^4)^{\frac{3}{4}}}{(0.5)^3)^{\frac{1}{3}}}

\\ \sf\longmapsto \dfrac{(0.15)^{2\times \dfrac{-3}{2}}\times (0.1)^{4\times \dfrac{3}{4}}}{(0.5)^{3\times \dfrac{1}{3}}}

\\ \sf\longmapsto \dfrac{(0.15)^{-3}(0.1)^3}{(0.5)^1}

\\ \sf\longmapsto \dfrac{\dfrac{1}{(0.15)^3}\times 0.001}{0.5}

\\ \sf\longmapsto \dfrac{\dfrac{1}{0.003375}(0.001)}{0.5}

\\ \sf\longmapsto\dfrac{ \dfrac{1000000}{3375}\times \dfrac{1}{1000}}{0.5}

\\ \sf\longmapsto \dfrac{\dfrac{1000}{3375}}{0.5}

\\ \sf\longmapsto \dfrac{1000}{3375}\times \dfrac{10}{5}

\\ \sf\longmapsto \dfrac{2000}{3375}

\\ \sf\longmapsto 0.592

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Set up an equation (part + part = whole). Solve the equation to find the value of x. 10 + x + x + 20 = 12
Licemer1 [7]

Answer:

x is equal to -9 thus it becomes -18 in the whole setup

3 0
3 years ago
Urgent help PLEASE NEED AWNSER
makkiz [27]
2s+s=18, i would say d but there are no letter options
3 0
3 years ago
Read 2 more answers
Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.
Jlenok [28]
y''+2y'+y=2t^2-t+3e^{-2t}

The characteristic equation is

r^2+2r+1=(r+1)^2=0\implies r=-1

so the characteristic solution is

y_c=C_1e^{-t}+C_2te^{-t}

For the particular solution, we can try looking for a solution of the form

y_p=at^2+bt+c+de^{-2t}
\implies{y_p}'=2at+b-2de^{-2t}
\implies{y_p}''=2a+4de^{-2t}

Substituting into the ODE, we have

(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}
at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}
\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3

So the general solution to the ODE is

y=y_c+y_p
y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}
4 0
3 years ago
) At a certain instant of time, a cube has side length 2 inches and its volume is increasing at a rate of 0.5 inches cubed per h
Phantasy [73]

Answer:

1 inches per cube

As required.

Step-by-step explanation:

Solution:

Given:

Volume = l3

dV / dt = 3 l2dl / dt

dv / dt = 0.5 inches cubed / hour

Surface area = 6 l2

Differentiate with respect to t: s = 6 l2

We get:

ds / dt = 12 l dl / dt

and dv / dt = = 0.5 inches cubed / hour

we know that:

dv / dt = 20 = 3 l2dl / dt

0.5 = 3 l2 (dl / dv)(dv / dt)

0.5 = 3l2 dl / dt

0.5 / 3l2 = dl / dt

Put 0.5 / 3l2 = dl / dt  in eq ds / dt = 12 l dl / dt

ds / dt = 12 l (0.5 / 3l2)

        =  4(0.5) / l

       = 2 / 2 inches pr cube    [ l = 2 inches per cube]

    = 1 inches per cube

As required.

3 0
3 years ago
Multiply using partial products
Alex
I don't see what we need to multiply but  here is an example: the numbers of the equation that I'm doing is  12 times 13 and this is how I do it .

  (1)(2)                           Step 1: multiply 3 times 2 then when you get the      
x 1 (3)                             answer you would need to multiply then 2 times 3.
  -------                                     
       (36) ( when multiplying 1 times 3 and then multiply 2 times 3)  
     
       
     Step 3: know that in the third step you would need to multiply 1 times 2 then 1 times 1.

    (1)(2)                                    
  x(1) 3
-------------                                        
       36
   + 120  (when multiplying 1 times 2 then 1 times 1 but I added a zero )
-----------
      156 ( I add them all up and that's the answer )

If need any questions just message me and I will answer back .




5 0
3 years ago
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