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3 years ago

Help pls :)) much appreciated

Mathematics

1 answer:

Ahat [919]3 years ago

6 0

Answer:

9,-7 hope this helps :))))))))

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. Two planes are 1620 miles apart and are traveling toward each other. One plane is traveling 120 mph

Nadusha1986 [10]

Answer:

Step-by-step explanation:

let the speed of one plane=x mph

speed of other plane=x+120 mph

x×1.5+(x+120)×1.5=1620

(x+x+120)×1.5=1620

(2x+120)×1.5=1620

3x+180=1620

3x=1620-180

3x=1440

x=480

speed of one plane=480 mph

speed of second plane=480+120=600 mph

4 0

3 years ago

Question 8, thanks guys

Natasha2012 [34]

Answer:

Step-by-step explanation:

Use the multiplication table of 12 and find that 5x12=60. Answer B is the only one that has that.

5 0

3 years ago

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.........................

maria [59]

Answer:

lol

Step-by-step explanation:

8 0

3 years ago

Please help with slope

ArbitrLikvidat [17]

Answer:

-2/3

Step-by-step explanation:

down 2, right 3

3 0

3 years ago

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EXAMPLE 7 Find the critical numbers of the function. f(x) = x3/5(8 − x). SOLUTION The Product Rule gives the following. f '(x) =

mash [69]

Answer:

Critical points: $x_{1} = 0$ and $x_{2} = 12$

Step-by-step explanation:

First, the function, which is a division of two functions, has to be derived:

$f'(x) = \frac{f(x)\cdot g'(x)-f'(x)\cdot g(x)}{[g(x)]^{2}}$ , where $f(x) = x^{3}$ and $g(x) = 5\cdot (8-x)$ .

The derivatives of each function are, respectively:

$f'(x) = 3\cdot x^{2}$

$g'(x) = -5$

All components are replaced and expressed is simplified afterwards:

$f'(x) = \frac{(x^{3})\cdot (-5)-(3\cdot x^{2})\cdot [5\cdot (8-x)]}{25}$

$f'(x) = \frac{-5\cdot x^{3}-120\cdot x^{2}+15\cdot x^{3}}{25}$

$f'(x) = \frac{10\cdot x^{3}-120\cdot x^{2}}{25}$

$f'(x) = \frac{2}{5}\cdot x^{3} - \frac{24}{5}\cdot x^{2}$

$f'(x) = \frac{2}{5}\cdot x^{2}\cdot(x-12)$

Let $f'(x) = 0$ be equalized to zero and critical numbers are found by the First Derivative Test:

$\frac{2}{5}\cdot x^{2}\cdot (x-12) = 0$

The critical points are:

$x_{1} = 0$ and $x_{2} = 12$

7 0

4 years ago