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Cloud [144]
3 years ago
13

Solve for m in D m/v, if D= 5.1 and v = 0.3 the / is for a fraction.

Mathematics
1 answer:
laila [671]3 years ago
8 0
5.1 = m/0.3
1.53 = m
Hope this helps!
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Find all the critical points of h(x) = x^3 − 3x^4 and categorize them as local
neonofarm [45]

Answer:

0 is an inflection point

1/4 is a local maximum.

Step-by-step explanation:

To begin with you find the first derivative of the function and get that

h'(x) = 3x^2 - 12x^3

to find the critical points you equal the first derivative to 0  and get that

3x^2 - 12x^3 = 0, x =  0,1/4

To find if they are maximums or local minimums you use the second derivative.

h''(x) = 6x-36x^2

since h''(0) = 0 is neither an inflection point, and since h''(1/4) = -3/4 then 1/4 is a maximum.

6 0
3 years ago
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rjkz [21]

Answer:

A

Step-by-step explanation:

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Minchanka [31]

Answer:

b. x ≥-2 or x ≤-6

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TIMED QUESTION NEED HELP FASTWhat values of b satisfy 3(2b + 3)2 = 36?
egoroff_w [7]

Answer:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

Step-by-step explanation:

3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12

Taking square root both sides, we get

\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}

Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

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3 years ago
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What is the index of the radical below? 4 radical 8

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