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pshichka [43]
3 years ago
5

Write this number in word form 125,210,000,

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
7 0

One hundred twenty-five million and two hundred ten thousand

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So the “certain number” will be S.
Since 2width + 2length, the first part would be 2 times 2 which is 4
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So it is 2(S-3) which is 2S-6
So the answer is 4+2S-6 which is 2S-2
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the cost of 3 pints of ice cream is 7 dollars . what is the proportion used to find the cost ,c, of 9 pints of the same ice crea
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Estimate then record the product 293*4
jolli1 [7]

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If f(x) = 3x - 5 and g(x) = 7x + 2, what is (f * g)(x)? NEED HELP ASAP!!!!
QveST [7]

Answer:(f*g)(x)= 21x^2 -29x -10


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21x^2-35x+6x-10

21x^2-29x-10

6 0
3 years ago
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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