Question

find a value of x other than 0 such that the vectors <-3x, 2x> and <4, x> are perpendicular.

Answer #1

When two vectors are perpendicular, then their dot product is zero.

or

or

That is

**Other value of x is 6**

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1.
a) Find a value of x other than 0 such that the vectors <-3x,
2x> and <4, x> are perpendicular
b) find the domain of the vector function r (t) = <sin (t),
ln (t), 1 / (t-1)>
c) Determine if the sequence converges or diverges, if it
converges determines its limit
ln (2 + e ^ n) / 2020n
d) Find the point (a, b, c) where the line x = 1-t, y = t, z = 1...

Consider the linearly independent set of vectors
B= (-1+2x+3x^2+4x^3+5x^4, 1-2x+3x^2+4x^3+5x^4,
1+2x-3x^2+4x^3+5x^4, 1+2x+3x^2-4x^3+5x^4, 1+2x+3x^3+4x^3-5x^4)
in P4(R), does B form a basis for P4(R) and why?

Find all the real roots of 6x^5 -x^4 + 2x^3 -3x^2 + 2x -18

f(x)=3x^4-7x^2+2x+1
Find the tangent line at x=2

find y' for the function
1. (y-2)^7=3x^2+2x-2
2. 3y^3+2x^3=3
3.(4y^2+3)^4+3x^5-5=0
4. 4x^2+3x^2y^2-y^3=3x

solve the initial value problem using Laplace
transform
x"(t)+3x'(t)+2x(t)=t
x(0)=0
x'(0)=2
differntial equations

Find (f −1 (a))' :
(a) f(x) = 2x 3 + 3x 2 + 7x + 4, a = 4
(b) f(x) = 2x + cos(x), a = 1

1. Find and interpret f'(−2)
f(x)=−2x^2−3x+4
f'(-2)=
2. Find and interpret f'(1)and f′'(1)
f(x)= 2/x
f'(1)=
f''(1)=
3. Find and interpret f'(1)and f′'(1)
f(x)=−2√x
f'(1)=
f''(1)=
4. If f(x)=(2x−5)(3x+1) then
f'(1)=
the values of x for which f(x)=0 are _ , _
the values of x for which f'(x)=0 are

Solve the following equations:
1) 4e^(2x)+5=17
2) e^(2x)-e^x-12=0
3) 2ln(x+1)=18
4) log(3x+2)-Log(4)=log(x+4)

solve the initial value problem: 2x-3e^(3x)y-e^(3x)y'=0
y(0)=2

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