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Scilla [17]
3 years ago
6

7 × (–3) × (–2)2 = ?

Mathematics
1 answer:
kiruha [24]3 years ago
7 0

\huge\text{Hey there!}

\huge\text{7*(-3) * (-2)(2) = ?}

\huge\text{7 * (-3) = -21}

\huge\text{-21 * (-2)2 = ?}

\huge\text{-21 * -2 = 42}

\huge\text{42(2) = ?}

\huge\text{42 * 2 = 42 + 42 = 84}

\boxed{\boxed{\huge\text{Answer: 84}}}\huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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(b) factorise <br>(3m-1)(6-a)-(m+3)(6-a)​
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Step-by-step explanation:

(3m-1)(6-a)-(m+3)(6-a)

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The differential equation y′′=0 has one of the following two parameter families as its general solution: yyyy=C1ex+C2e−x=C1cos(x
svet-max [94.6K]

Answer:

y_c = 2 + 10*x

Step-by-step explanation:

Given:

                                                y'' = 0

Find:

- The solution to ODE such that y(0) = 2, y'(0) = 10

Solution:

- Assuming a solution y = Ce^(mt)

So,                                y' = C*me^(mt)

                                    y'' = C*m^2e^(mt)

- Back substitute into given ODE, we get:

                                    y'' = C*m^2e^(mt) = 0

                                    e^(mt) can not be equal to zero

- Hence,                       m^2 = 0

                                     m = 0 , 0 - (repeated roots)

- The complimentary function for repeated roots is:

                                    y_c = (C1 + C2*x)*e^(m*t)

                                    y_c = C1 + C2*x  

- Evaluate @ y(0) = 2

                                    2 = C1 + C2*0

                                    C1 = 2

-Evaluate @ y'(0) = 10

                                    y'(t) = C2 = 10

Hence,                         y_c = 2 + 10*x

5 0
3 years ago
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