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3 years ago

The graph shows the distances traveled by two toy trains.

Mathematics

2 answers:

liraira [26]3 years ago

4 0

From the graph ...

-- Train-1 travels more than 15 feet in 3 seconds.

-- Train-2 travels less than 15 feet in 3 seconds.

So, if Train-3 travels exactly 15 feet in 3 seconds,
then Train-3 is faster than #2 but slower than #1 .

Ivenika [448]3 years ago

3 0

Answer:D. The third train is faster than Train 2 but slower than Train 1.

Step-by-step explanation:

Since, If the speed of one object is greater than that of other object then we can say that one object is faster than other object.

Here, the speed of third train = 15/3=5 feet/sec ( because speed = distance/time)

And, according to the given graph the speed of train 1= 30/5=6 feet/sec ( by taking the approximate value of x and y by the graph)

While, the speed of train 2= 20/5= 4 feet/sec

Thus, we can say that, Speed of train 1>Speed of train 3> Speed of train 2

Therefore, Only Option D is correct.

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Find all solutions to x³+3x² − 9x − 27 = 0 by factoring the equation.

SCORPION-xisa [38]

Answer:

The solutions are $x=-3,\:x=3$

Step-by-step explanation:

To factor this cubic polynomial $x^3+3x^2-9x-27$ you must:

Group the polynomial into two sections

$x^3+3x^2-9x-27=\left(x^3+3x^2\right)+\left(-9x-27\right)$

Factor out -9 from $(-9x-27)$

$(-9x-27)=-9\left(x+3\right)$

Factor out $x^2$ from $(x^3+3x^2)$

$(x^3+3x^2)=x^2\left(x+3\right)$

$x^3+3x^2-9x-27=-9\left(x+3\right)+x^2\left(x+3\right)$

Factor out common term $x+3$

$-9\left(x+3\right)+x^2\left(x+3\right)=\left(x+3\right)\left(x^2-9\right)$

$x^3+3x^2-9x-27=\left(x+3\right)\left(x^2-9\right)$

Factor $x^2-9$

$x^2-9=\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27= \left(x+3\right)\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27=\left(x+3\right)^2\left(x-3\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+3=0, \quad{x=-3}\\x-3=0, \quad{x=3}$

The solutions are

$x=-3,\:x=3$

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2 years ago

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