Answer:
Less than
Step-by-step explanation:
6m3 - 16m2 + 15m - 40<span> Simplify —————————————————————
2m2 + 5
</span>Checking for a perfect cube :
<span> 4.1 </span> <span> 6m3 - 16m2 + 15m - 40</span> is not a perfect cube
Trying to factor by pulling out :
<span> 4.2 </span> Factoring: <span> 6m3 - 16m2 + 15m - 40</span>
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 15m - 40
Group 2: <span> -16m2 + 6m3</span>
Pull out from each group separately :
Group 1: (3m - 8) • (5)
Group 2: <span> (3m - 8) • (2m2)</span>
-------------------
Add up the two groups :
<span> (3m - 8) • </span><span> (2m2 + 5)</span>
<span>Which is the desired factorization</span>
<span>3m-8 is the answer</span>
Answer:
Except 4 out of 6
the answer is the third one
Simplify
9x^3 - 4x + 10 + x^3 + 10x - 9
Collect like terms
(9x^3 + x^3) + (-4x + 10x) + (10 - 9)
Simplify
<u>10x^2 + 6x + 1</u>
<h3><u>Solution</u></h3>
<u>Given </u><u>:</u><u>-</u>
- Perimeter of rectangle = 72 cm
- The length is 3 more than twice the width.
<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>:</u><u>-</u>
<h3 /><h3>
<u>Explantion</u></h3>
<u>Using </u><u>Formula</u>

<u>Let,</u>
- Length of Rectangle = x cm
- Breadth of Rectangle = y cm
<u>According</u><u> to</u><u> question</u><u>,</u>
==> perimeter of Rectangle = 72
==> 2(x+y) = 72
==> x + y = 72/2
==> x + y = 36_________________(1)
<u>Again,</u>
==> x = 2y + 3
==> x - 2y = 3__________________(2)
<u>Subtract</u><u> </u><u>equ(</u><u>1</u><u>)</u><u> </u><u>&</u><u> </u><u>equ(</u><u>2</u><u>)</u>
==> y + 2y = 36 - 3
==> 3y = 33
==> y = 33/3
==> y = 11
<u>keep </u><u>in </u><u>equ(</u><u>1</u><u>)</u>
==> x - 2×11 = 3
==> x = 3 + 22
==> x = 25
<h3><u>Hence</u></h3>
- <u>Length</u><u> of</u><u> </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u> </u><u>cm</u>
- <u>Width </u><u>of </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u> </u><u>cm</u>
<h3>
<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h3>
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