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3 years ago

Factor x3 – 7x2 – 5x + 35 by grouping. What is the resulting expression?

Mathematics

1 answer:

zmey [24]3 years ago

4 0

[ Answer ]

$\boxed{-16x \ + \ 35}$

[ Explanation ]

Factor x3 - 7 · 2 - 5x + 25 By Grouping

--------------------------------------------

Add Similar Elements: 3x - 5x = -2x

-2x - 7 · 2x + 35

Multiply Numbers: 7 · 2 = 14

-2x - 14x + 35

Add Similar Elements: -2x - 14x = -16x

-16x + 35

$\boxed{[ \ Eclipsed \ ]}$

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An attacker at the base of a castle wall 4 m high throws a rock straight up with speed 7.5 m/s from a height of 1.5 m

Sergeu [11.5K]

Answer:

a) Yes, b) $v\approx 2.686\,\frac{m}{s}$ , c) $v_{o} \approx 7.002\,\frac{m}{s}$

Step-by-step explanation:

a) The maximum height reached by the rock is:

$\Delta y = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(7.5\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 2.868\,m$

$y = 1.5\,m + 2.868\,m$

$y = 4.368\,m$

Yes, the rock will exceed the top of the wall.

b) The speed when the rock reaches the top of the wall:

$v = \sqrt{\left(7.5\,\frac{m}{s} \right)^{2}+2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m-1.5\,m)}$

$v\approx 2.686\,\frac{m}{s}$

c) The initial speed required so that stone does not exceed the top of the wall is:

$v_{o} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m-1.5\,m) }$

$v_{o} \approx 7.002\,\frac{m}{s}$

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4 years ago

Subtract.

faltersainse [42]

18 1/2 -17 3/4
= 18.50 -17.75
= 0.75
= 3/4

The 4th selection is appropriate.

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3 years ago

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Vadim26 [7]

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Step-by-step explanation:

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