Apply to Row 2 : Row 2 + Row 1
2x + 2y + 3z = 0
y + 4z = -3
2x + 3y + 3z = 5
Apply to Row 3: Row 3 - Row 1
2x + 2y + 3z = 0
y + 4z = -3
y = 5
Apply to Row 3: Row 3 - Row 2
2x + 2y + 3z = 0
y + 4z = -3
-4z = 8
Simplify rows
2x + 2y + 3z = 0
y + 4z = -3
z = -2
<em>Note that the matrix is in echelon form now. The next steps are for back substitution.</em>
Apply to Row 2: Row 2 - 4 Row 3
2x + 2y + 3z = 0
y = 5
z = -2
Apply to Row 1: Row 1 - 3 Row 3
2x + 2y = 6
y = 5
z = -2
Apply to Row 1: Row 1 - 2 Row 2
2x = -4
y = 5
z = 2
Simplify the rows
<u>x = -2</u>
<u>y = 5</u>
<u>z = -2</u>
2x^2 because you do (-4x^2)-x=(-6x^2) and solve for x to get 2x^2.
![\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x} \\\\\\ (-2)^3=x\implies \boxed{-8=x}\\\\ -------------------------------\\\\ 0=\sqrt[3]{x}\implies \boxed{0=x}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dx%2B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B3%5Cleft%28%5Cfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%20%20%5Cright%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B%5Ccfrac%7B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Csqrt%5B3%5D%7Bx%7D%2B2%5Cimplies%20-2%3D%5Csqrt%5B3%5D%7Bx%7D%0A%5C%5C%5C%5C%5C%5C%0A%28-2%29%5E3%3Dx%5Cimplies%20%5Cboxed%7B-8%3Dx%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A0%3D%5Csqrt%5B3%5D%7Bx%7D%5Cimplies%20%5Cboxed%7B0%3Dx%7D)
now, f(0) = 0, and f(-8) is an imaginary value or no real value.
now, f(-10) will also give us an imaginary value
and f(1) = 4
so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.
however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.
the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).
Answer:
88 feet per second
Step-by-step explanation:
quick maths
Which is an infinite arithmetic sequence? a{10, 30, 90, 270, …} b{100, 200, 300, 400} c{150, 300, 450, 600, …} d{1, 2, 4, 8}
umka21 [38]
Answer:
C
Step-by-step explanation:
An arithmetic sequence has a common difference d between consecutive terms.
Sequence a
30 - 10 = 20
90 - 30 = 60
270 - 90 = 180
This sequence is not arithmetic
Sequence b
200 - 100 = 100
300 - 200 = 100
400 - 300 = 100
This sequence is arithmetic but is finite, that is last term is 400
Sequence c
300 - 150 = 150
450 - 300 = 150
600 - 450 = 150
This sequence is arithmetic and infinite, indicated by ........ within set
Sequence d
2 - 1 = 1
4 - 2 = 2
8 - 4 = 4
This sequence is not arithmetic
Thus the infinite arithmetic sequence is sequence c
