F '(x<span>) = </span>2(3x2<span> + 5) + 6x(</span>2x<span> - </span>1<span>) ... + </span>3), s(x<span>) = </span>x<span> - 4. Use the </span>product<span> rule table. below to get: r'(</span>x) = 20x + 11. s'(x<span>) = </span>1<span> ... f'(</span>x) = (x<span> - 4</span>)(20x + 11) - (5x<span> - </span>2)(2x<span> + </span>3)/(x<span> - 4)</span>2<span>.</span>First, we will distribute 2x<span> to (</span>x<span> + 5), then we will distribute </span>3<span>. ... </span>5x<span>. That is,. multiplying binomials. 6x − </span>x<span>= </span>5x<span>. Example </span>2<span>. Multiply (3x − </span>1)(x<span> + </span>2). Answer. 3x2<span> + </span>5x<span>
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Answer:
Joe worked 3 hours overtime.
Step-by-step explanation:
Let t = number of hours overworked
425 + 20 × t = 485
425 + 20t = 485
20t = 485 - 425
= 60
t = 3
Joe worked 3 hours over time.
Hope this helps :)
180 is that I think not for sure
Answer:
x^2+0x+(-x^2)=0
Explanation:
x^2+0x+(-x^2)=0
Anything multiplied by 0 is equal to 0
x^2+0+(-x^2)=0
x^2+(-x^2)=0
A positive and a negative makes a negative
x^2-x^2=0
0=0
For these types of problems, play around with multiplying by 0 and adding negatives.
I hope this helps! Please comment if you have any questions.
X to the one half power, over x to the three eighteenth power is equal to x to the one half power, divided by x to the one sixth power, which equals x to the power of (one half minus one sixth), or x to the one third power.
<span>The twenty seventh root of the quantity of x to the second times x to the third times x to the fourth equals the twenty seventh root of x to the ninth power which equals x to the one third power. </span>
<span>I cannot say whether Francisco and Ryan started with equivalent expressions but on final simplification they ended up the same. One might reasonably assume they started with equivalent expressions, but who knows if they made any mistakes in their simplifications.</span>
