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Olenka [21]
4 years ago
13

How do I show my work for this ?

Mathematics
1 answer:
Zarrin [17]4 years ago
3 0
First, you multiply all terms by 2.8 to get rid of the denominator:
m-13.72 = -19.908
then, you isolate the m on one side of the equation and all other terms on the other side:
m = -19.908+13.72
m = -6.188
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rectangular solid that is 6-cm-by-6 cm-by-6 cm is painted on all six faces. then the solid is cut into cubes that measure 2 cm o
Nitella [24]
If you check the picture below

the rectangular solid is a 6x6x6, so all sides are equal thus is a cube

is painted on all faces, so the only part that's painted is the faces, ok
then you split it like in the picture, you have all 6 faces painted there,

now, let's say each of those 6 faces are cut in 2cm

how many of those 2cm pieces will each have?   so get that amount for each and sum them up, that's how many of the cubes have one face painted

only the cubes that are part of the face, which was painted, are the ones that have a painted face themselves

keep in mind that each face is a 6x6 rectangle
notice the 2nd picture below

6 0
3 years ago
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JulsSmile [24]

Answer: its b ;p

Step-by-step explanation:

4 0
4 years ago
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I need this answered its due soon
mixer [17]
<h2><u>C D E</u></h2>

A is false because this graph doesn't have any relative minimums because it never increases

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7 0
3 years ago
Solve y = x^2 +11 <br> for x.
lisabon 2012 [21]

Answer:

x = ± \sqrt{y-11}

Step-by-step explanation:

Given

y = x² + 11 ( subtract 11 from both sides )

y - 11 = x² ( take the square root of both sides )

± \sqrt{y-11} = \sqrt{x^2}, hence

x = ± \sqrt{y-11}

7 0
4 years ago
Read 2 more answers
Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.
igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

where m₁ = 3 and m₂ = -7.

So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

6 0
4 years ago
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