MATHEMATICS COLLEGE

Which way is counterclockwise?A. The way the hands move on a clock. (To the right)
B. The opposite way the hands move on a clock. (To the left)

Answers

Answer 1
Answer:

Answer:

B. the opposite way the hands move on a clock

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I am probably just really dumb, but I don't understand the concept. The question is: -42v+33<8v+91. I am supposed to solve the inequality for v. Please give an explanation and not just an answer, I would really like to know how to do this.

Answers

-42v +33 <8v +91

Move all terms with V to the left side by subtracting 8V from both sides.

-50v +33 < 91

Now get the term with V all by itself by subtracting 33 from each side:

-50v<58

Solve for V by dividing both sides by -50  ( when dividing an inequality by a negative number you also need to reverse the direction of the inequality sign).

v >58/-50

Reduce the fraction and relocate the negative sign:

V >-29/25

Coeffitiant of 3x-y+4

Answers

Answer:

3

Step-by-step explanation:

....,.................
Your answer Would be A

-6(x - 6) = x(16 - 7)
Help please

Answers

Answer:

the answer would be x equal 2.4

Step-by-step explanation:

sorry for my bad handwriting

A(2,9), B(4,k), and C(9, -12) are 3 collinear points.
Find the value of k.

Answers

Answer is   3

==========================================================

Explanation:

We're going to be using the slope formula a bunch of times.

Find the slope of the line through points A and C

m = (y2 - y1)/(x2 - x1)

m = (-12-9)/(9-2)

m = -21/7

m = -3

The slope of line AC is -3. The slopes of line AB and line BC must also be the same for points A,B,C to be collinear. The term collinear means all three points fall on the same straight line.

-------

Let's find the expression for the slope of line AB in terms of k

m = (y2 - y1)/(x2 - x1)

m = (k-9)/(4-2)

m = (k-9)/2

Set this equal to the desired slope -3 and solve for k

(k-9)/2 = -3

k-9 = 2*(-3) ..... multiply both sides by 2

k-9 = -6

k = -6+9 .... add 9 to both sides

k = 3

If k = 3, then B(4,k) updates to B(4,3)

-------

Let's find the slope of the line through A(2,9) and B(4,3)

m = (y2 - y1)/(x2 - x1)

m = (3-9)/(4-2)

m = -6/2

m = -3 we get the proper slope value

Finally let's check to see if line BC also has slope -3

m = (y2 - y1)/(x2 - x1)

m = (-12-3)/(9-4)

m = -15/5

m = -3 we get the same value as well

Since we have found lines AB, BC and AC all have slope -3, we have proven that A,B,C fall on the same straight line. Therefore, this shows A,B,C are collinear.

Find the 1000th term for the sequence

Answers

Answer:

D. 7017

Step-by-step explanation:

if 24 is the first term, find 7x999, or 7x1000-7 and add 24

however a better way would be to use the formula

value=a+(n-1)d

a = the first term in the sequence (24)

n     =     the amount of terms you need (1000)

d = the common difference between terms (7)

Solve this differential Equation by using power series
y''-x^2y=o

Answers

We're looking for a solution

y=\displaystyle\sum_(n=0)^\infty a_nx^n

which has second derivative

y''=\displaystyle\sum_(n=2)^\infty n(n-1)a_nx^(n-2)=\sum_(n=0)^\infty(n+2)(n+1)a_(n+2)x^n

Substituting these into the ODE gives

\displaystyle\sum_(n=0)^\infty(n+2)(n+1)a_(n+2)x^n-\sum_(n=0)^\infty a_nx^(n+2)=0

\displaystyle\sum_(n=0)^\infty(n+2)(n+1)a_(n+2)x^n-\sum_(n=2)^\infty a_(n-2)x^n=0

\displaystyle2a_2+6a_3x+\sum_(n=2)^\infty(n+2)(n+1)a_(n+2)x^n-\sum_(n=2)^\infty a_(n-2)x^n=0

\displaystyle2a_2+6a_3x+\sum_(n=2)^\infty\bigg((n+2)(n+1)a_(n+2)-a_(n-2)\bigg)x^n=0

Right away we see a_2=a_3=0, and the coefficients are given according to the recurrence

\begin{cases}a_0=y(0)\na_1=y'(0)\na_2=0\na_3=0\nn(n-1)a_n=a_(n-4)&\text{for }n\ge4\end{cases}

There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.

  • If n=4k, where k\ge0 is an integer, then

k=0\implies n=0\implies a_0=a_0

k=1\implies n=4\implies a_4=(a_0)/(4\cdot3)=\frac2{4!}a_0

k=2\implies n=8\implies a_8=(a_4)/(8\cdot7)=(6\cdot5\cdot2)/(8!)a_0

k=3\implies n=12\implies a_(12)=(a_8)/(12\cdot11)=(10\cdot9\cdot6\cdot5\cdot2)/(12!)a_0

and so on, with the general pattern

a_(4k)=(a_0)/((4k)!)\displaystyle\prod_(i=1)^k(4i-2)(4i-3)

  • If n=4k+1, then

k=0\implies n=1\implies a_1=a_1

k=1\implies n=5\implies a_5=(a_1)/(5\cdot4)=(3\cdot2)/(5!)a_1

k=2\implies n=9\implies a_9=(a_5)/(9\cdot8)=(7\cdot6\cdot3\cdot2)/(9!)a_1

k=3\implies n=13\implies a_(13)=(a_9)/(13\cdot12)=(11\cdot10\cdot7\cdot6\cdot3\cdot2)/(13!)a_1

and so on, with

a_(4k+1)=(a_1)/((4k+1)!)\displaystyle\prod_(i=1)^k(4i-1)(4i-2)

  • If n=4k+2 or n=4k+3, then

a_2=0\implies a_6=a_(10)=\cdots=a_(4k+2)=0

a_3=0\implies a_7=a_(11)=\cdots=a_(4k+3)=0

Then the solution to this ODE is

\boxed{y(x)=\displaystyle\sum_(k=0)^\infty a_(4k)x^(4k)+\sum_(k=0)^\infty a_(4k+1)x^(4k+1)}
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