If you feel better at school, then you will eat breakfast.
A.)
x y
1 1
2 4
3 9
dx=2-1→dx=1 dy=4-1→dy=3
dx=3-2→dx=1 dy=9-4→dy=5 different to 3, then this table can not be represented by a line
B.)
x y
1 2
2 5
3 10
dx=2-1→dx=1 dy=5-2→dy=3
dx=3-2→dx=1 dy=10-5→dy=5 different to 3, then this table can not be represented by a line
C.)
x y
1 3
2 6
3 9
dx=2-1→dx=1 dy=6-3→dy=3
dx=3-2→dx=1 dy=9-6→dy=3, then this table can be represented by a line
D.)
x y
1 0
2 3
3 8
dx=2-1→dx=1 dy=3-0→dy=3
dx=3-2→dx=1 dy=8-3→dy=5 different to 3, then this table can not be represented by a line
Answer: Option C. x 1 2 3 y 3 6 9
Hello from MrBillDoesMath!
Answer:
(x^4-2) (x^4 -1)
Discussion:
Let u = x^4, then
x^8 - 3x^4 + 2
= u^2 - 3u + 2
= (u -2) ( u -1)
= (x^4-2) (x^4 -1)
Thank you,
MrB
Answer:
Step-by-step explanation:
To multiply radicals with the same root, you just need to multiply the numbers or expressions inside the root retaining the same root and then simplify the result.
In this case: 
(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.
