Find the difference (16x^3 + 2x^2) - 4x^3 =

Mathematics

1 answer:

djverab [1.8K]4 years ago

8 0

I got 2x^2•(6x + 1) hope this helps :)

You might be interested in

Convert the following sum into a product. cosx + cos3x + cos5x + cos7x

kvv77 [185]

Answer:

$4\cos x\cos (4x)\cos (2x)$

Step-by-step explanation:

Given: $cosx + cos3x + cos5x + cos7x$

To convert: the given sum into product

Solution:

Use formula: $\cos A+\cos B=2\cos \left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos (2x)\cos (-x)+2\cos (6x)\cos (-x)\\=2\cos (2x)\cos (x)+2\cos (6x)\cos (x)\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]\\=2\cos x\left [2 \cos \left ( \frac{2x+6x}{2} \right )\cos \left ( \frac{2x-6x}{2} \right ) \right ]\\=2\cos x\left [ 2\cos (4x) \cos (-2x) \right ]\\=4\cos x\cos (4x)\cos (2x)$

4 0

3 years ago

Need answers ASAP What is m∠LQK ? Enter your answer in the box.

mojhsa [17]

It is an acute angle.

8 0

3 years ago

Read 2 more answers

Graph the following features: Slope = -7/5 Y-intercept = 1

valentinak56 [21]

Answer:

Step-by-step explanation:

1) in order to graph the required direct line it needed two points:

2) one of these two points is Y-intercept. According to the condition it is (0;1);

3) in order to find the second point, it is enough to apply the given slope: to start in point (0;1), then to move to the left direction (-5 units, move no. 1), then to move up (+7 units, move no. 2);

4) to graph the line between two points (red line); y= -7/5x +1.

All the details are in the attachment.