Answer: 3^12 x 7^9 = 2.144556126 x 10^13
Step-by-step explanation:
3^12 = 531,441
7^9 = 40,353,607
531,441 x 40,353,607 = 2.144556126 x 10^13
So distribute using distributive property
a(b+c)=ab+ac so
split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24
the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24
Answer:
Problem 20)
Problem 21)
A)
The velocity function is:
The acceleration function is:
B)
Step-by-step explanation:
Problem 20)
We want to differentiate the equation:
We can take the natural log of both sides. This yields:
Since ln(aᵇ) = bln(a):
Take the derivative of both sides with respect to <em>x: </em>
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Implicitly differentiate the left and use the product rule on the right. Therefore:
Simplify:
Simplify and multiply both sides by <em>y: </em>
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Since <em>y</em> = (cos x)ˣ:
Problem 21)
We are given the position function of a particle:
A)
Recall that the velocity function is the derivative of the position function. Hence:
Differentiate:
The acceleration function is the derivative of the velocity function. Hence:
Differentiate:
B)
The position at <em>t</em> = 0 will be:
The velocity at <em>t</em> = 0 will be:
And the acceleration at <em>t</em> = 0 will be: