Is the algebra, if so what type?
Answer:
no 15x — 20xy as 5x(3— 4xy is not correct
Step-by-step explanation:
15x — 20xy ≠ 5x(3— 4xy)
there are not equal because 5x(3 - 4xy) =
when you simplify the bracket, you will have
15x - 20x²y
Answer:
Self selection sampling.
Step-by-step explanation:
A poll that does not attempt to generate a random sample, but instead invites people to volunteer to participate is called - self selection sampling.
Self-selection sampling is a sampling method where researchers allow the people or individuals, to choose to take part in research on their own accord.
Answer: No, he can't pass the course.
Explanation:
Since we have given that
Marks on tests 1, 2, and 4 are given by
82, 76, 90 respectively.
Unfortunately, he cut the third test and receive a 0.
According to question, we have given that passing grade for the course is 70, and his one test left to take ,
Let the marks obtained in fifth test be x
So,

And it is not possible to get 102 over 100.
so, he can't still pass the course.
Hence, No, he can't pass the course.
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.