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mezya [45]
3 years ago
7

Consider the system of equations. which shows an equivalent system of equations? see attatched

Mathematics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

A.

Step-by-step explanation:

All choices show the first original equation.

Each choice shows a different second equation.

Take the second original equation:

6x + 5y = 30

Multiply both sides by -5.

-30x - 25y = -150

Answer: A.

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How long is a line of 1 million centimeter cubes be?
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Evaluate the expression below using the properties of operations. Show your steps.−36 ÷ 14 ⋅ (−18) ⋅ (−3) ÷ 6
s2008m [1.1K]

Answer:

(-162)/7 or -23 1/7 as mixed fraction

Step-by-step explanation:

Simplify the following:

(-36)/14 (-18) (-3)/6

Hint: | Express (-36)/14 (-18) (-3)/6 as a single fraction.

(-36)/14 (-18) (-3)/6 = (-36 (-18) (-3))/(14×6):

(-36 (-18) (-3))/(14×6)

Hint: | In (-36 (-18) (-3))/(14×6), divide -18 in the numerator by 6 in the denominator.

(-18)/6 = (6 (-3))/6 = -3:

(-36-3 (-3))/14

Hint: | In (-36 (-3) (-3))/14, the numbers -36 in the numerator and 14 in the denominator have gcd greater than one.

The gcd of -36 and 14 is 2, so (-36 (-3) (-3))/14 = ((2 (-18)) (-3) (-3))/(2×7) = 2/2×(-18 (-3) (-3))/7 = (-18 (-3) (-3))/7:

(-18 (-3) (-3))/7

Hint: | Multiply -18 and -3 together.

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(54 (-3))/7

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Answer:  (-162)/7

3 0
3 years ago
Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas
o-na [289]
Answers:

(a) f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing at (-2,2).

(c) f is concave up at (2, \infty)

(d) f is concave down at (-\infty, 2)

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

f'(x) = 15x^2 - 60


So,


f'(x) \ \textgreater \  0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \  0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \  0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \  0} \text{   (1)}

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
 If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

f'(x) \ \textgreater \  0 \Leftrightarrow (x - 2)(x + 2)  \ \textgreater \  0

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing only when the derivative of f is negative. Since

f'(x) = 15x^2 - 60

Using the similar computation in (a), 

f'(x) \ \textless \  \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \  0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \  0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \  0} \text{ (2)}

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

f''(x) = 30x - 60

Since,

f''(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30x - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30(x - 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow x - 2 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow \boxed{x \ \textgreater \  2}

Therefore, f is concave up at (2, \infty).

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

f''(x) = 30x - 60

Using the similar computation in (c), 

f''(x) \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30x - 60 \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30(x - 2) \ \textless \  0 &#10;\\ \\ \Leftrightarrow x - 2 \ \textless \  0 &#10;\\ \\ \Leftrightarrow \boxed{x \ \textless \  2}

Therefore, f is concave down at (-\infty, 2).
3 0
3 years ago
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