Max value of x is infinite and min value of x is 0
Answer:
it is on the nearest tenth. The number on the other side of the dot are tenths
Bruh. It's obviously 4. That's the only one that fits into 28 evenly.
Answer:
Step-by-step explanation:
Look at both terms separately, and isolate each "part."
i.e. 12x^4y^2 becomes 12, x^4, y^2
So the problem becomes:
12, x^4, y^2 and
30, x^3, y^1
Then, find the GCF of each term,
30 and 12 - 6
x^4 and x^3 - x^3
y^2 and y^1 - y^1
Hope it helps :) and let me know if you want me to elaborate.
The solution to the problem is as follows:
let y = asinx + bcosx
<span>
dy/dx = acosx - bsinx </span>
<span>
= 0 for max/min </span>
<span>
bsinx = acosx </span>
<span>
sinx/cosx = a/b </span>
<span>
tanx = a/b </span>
<span>
then the hypotenuse of the corresponding right-angled triangle is √(a^2 + b^2) </span>
<span>the max/min of y occurs when tanx = a/b </span>
<span>
then sinx = a/√(a^2 + b^2) and cosx = b/√(a^2 + b^2) </span>
<span>
y = a( a/√(a^2 + b^2)) + b( b/√(a^2 + b^2)) </span>
<span>
= (a^2 + b^2)/√(a^2 + b^2) </span>
<span>
= √(a^2 + b^2)</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
