Find the GCF of the terms of the polynomial. 33x^3 + 18^6

1 answer:

5 0

$33x^3=\boxed{3}\cdot11\cdot\boxed{x}\cdot\boxed{x}\cdot\boxed{x}\\\\18x^6=2\cdot\boxed{3}\cdot3\cdot\boxed{x}\cdot\boxed{x}\cdot\boxed{x}\cdot x\cdot x\cdot x\\\\GCF(33x^3,\ 18x^6)=\boxed{3}\cdot\boxed{x}\cdot\boxed{x}\cdot\boxed{x}=3x^3$

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Help please with b. and d.

cupoosta [38]

I'm assuming the question is asking for you to factor the expressions.

b.
(x-3)(x+2)
-3+2=-1
-3*2=-6

d.
(x-6)(x-7)
-6-7=-13
-6*-7=42

3 0

3 years ago

How do you solve 1 5/6• 2 2/9 ?

nevsk [136]

The answer to your question is 2 5/27

5 0

3 years ago

Read 2 more answers

A particle is moving along the curve y= 4sqrt(5x+11) . As the particle passes through the point (5,24) , its -coordinate increas

kupik [55]

The rate of change of the distance from the particle to the origin at this instant is 3 units per second.

<h3>What is the rate of change?</h3>

The instantaneous rate of change is the rate of change at a particular instant.

A particle is moving along the curve

$y= 4\sqrt{5x+11} .$

The rate of change of y is given as:

dy / dx = 2

by differentiating both sides,

$\dfrac{dy}{dx} = 4\dfrac{1}{2\sqrt{5x+11} } 5\dfrac{dx}{dt} \\\\\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\$

From the question, we have:

(x, y) = (5,24)

Substitute 5 for x and dy / dx = 2

$\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\\\5= \dfrac{10}{\sqrt{5(5)+11} }\dfrac{dx}{dt}\\\\\\\sqrt{5(5)+11} = 2\dfrac{dx}{dt}\\\\\\\dfrac{dx}{dt} = 6/ 2 = 3$