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Lunna [17]
3 years ago
15

How do you find the surface area of a 3d object????

Mathematics
2 answers:
Igoryamba3 years ago
8 0
You should start by calculating the area of one of the faces of the chosen figure. Then proceed by doing the same thing till you've calculated the area of all the surfaces. Then all you have to do is add them up.
olga2289 [7]3 years ago
4 0
If you have enough info google can do all the work for you.
You might be interested in
Please help me to find value P ​
PSYCHO15rus [73]
P=10
Explanation
2p+p+90+100+140=360
3p+330=360
-330. -330
3p=30
P=10
8 0
3 years ago
Find the value of $x$ that maximizes $$f(x) = \log (-20x + 12\sqrt{x}).$$ If there is no maximum value, write "NONE".
kakasveta [241]
The function is written as:
f(x) = log(-20x + 12√x)
To find the maximum value, differentiate the equation in terms of x, then equate it to zero. The solution is as follows.

The formula for differentiation would be:
d(log u)/dx = du/u ln(10)
Thus,
d/dx = (-20 + 6/√x)/(-20x + 12√x)(ln 10) = 0
-20 + 6/√x = 0
6/√x = 20
x = (6/20)² = 9/100

Thus,
f(x) =  log(-20(9/100)+ 12√(9/100)) = 0.2553

<em>The maximum value of the function is 0.2553.</em>
5 0
3 years ago
PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
umka2103 [35]
That kinda look like a roof from fortnite
6 0
3 years ago
Read 2 more answers
Use the variation of parameters method to solve the DR y" + y' - 2y = 1
postnew [5]

Answer:

y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

y"+y'-2y=1

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

D^2+D-2\ =\ 0

=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}

=>\ D=\ -2\ or\ 1

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

y_c(t)\ =\ C_1e^{-2t}+C_2e^t

Let's assume that

     y_1(t)=e^{-2t}          y_2(t)=e^t

=>\ y'_1(t)=-2e^{-2t}        y'_2(t)=e^t

   and g(t)=1

Now, the Wronskian can be given by

W=y_1(t).y'_2(t)-y'_1(t).y_2(t)

   =e^{-2t}.e^t-e^t(-e^{-2t})

   =e^{-t}+2e^{-t}

   =3e^{-t}

Now, the particular solution can be given by

y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}

=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}

=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}

=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}

=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

Now, the complete solution of the given differential equation can be given by

y(t)\ =\ y_c(t)+y_p(t)

      =C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

5 0
3 years ago
Determine whether a figure with the given vertices is a parallelogram. A(4,5), B(-7,-10), C(-4,-9), D(7,6) distance and slope fo
Ymorist [56]

Answer:

C i think..

Step-by-step explanation:

8 0
3 years ago
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