MATHEMATICS COLLEGE

Subtract the following military time: 0545 - 0058

Answers

Answer 1

Answer: You subtract it and get 487

Answer 2

Answer:

Answer:

i think it would be 0447

Step-by-step explanation:

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Consider w1 = 4 + 2i and w2 = –1 – 3i. Which graph represents the sum w1 + w2?
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If x3 = 8 and y3 = 125, whatis the value of y - x?

I need these please

Answers

Answer:

Step-by-step explanation:

2x+30=180

2x = 150

x = 75

y=30

Help!!!

prove triangle PAB is an equilateral triangle

HELP PLEASEE

Answers

Answer:

If three sides of a triangle are equal and the measure of all three angles is equal to 60 degrees then the triangle is an equilateral triangle. Therefore it is an equilateral triangle.

In ∆PAM and ∆PBM

$\because\pink{\sf\begin{cases} \sf AM=BM(Bisected\:by\:PM)\n \sf PA=PB(Given)\n \sf PM(Common\:side)\end{cases}}$

$\n \sf\longmapsto ∆PAM\cong ∆PBM(SSS)$

$\n \sf\longmapsto PA+AM=PB+BM$

Put AM =BM

$\n \sf\longmapsto PA+AM+AM=PB+BM+BM$

Put PA=PB

$\n \sf\longmapsto PA+AB=PB+AB$

Hence

∆APB is a equilateral triangle

The difference between a number x and 12 is 16?

Answers

Answer:

x equals 28

Step-by-step explanation:

16 + 12 = x because

This is the two that go together, but are separate questions

Answers

Answers:

Option 1)

6a + 8s = 102

14a + 4s = 150

each adult ticket costs 9 dollars

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

Explanation:

6 adult tickets and 8 student tickets bring in $102, so that means 6a+8s = 102

14 adult tickets and 4 students tickets bring in $150, so 14a+4s = 150

The system of equations is

$\begin{cases}6a+8s = 102\n14a+4s = 150\end{cases}$

If we multiply both sides of the second equation by -2, we get this updated system

$\begin{cases}6a+8s = 102\n-28a-8s = -300\end{cases}$

Add the equations straight down

6a+(-28a) = -22a

8s+(-8s) = 0s = 0 ... the 's' terms go away

102+(-300) = -198

So we end up with the equation -22a = -198 and that solves to a = 9 after dividing both sides by -22.

Each adult ticket costs $9

If you want the value of s, then

6a+8s = 102

6(9)+8s = 102

54 + 8s = 102

8s = 102-54

8s = 48

s = 48/8

s = 6

Meaning each student ticket costs $6

Or you could use the other equation

14a+4s = 150

14(9)+4s = 150

126+4s = 150

4s = 150-126

4s = 24

s = 24/4

s = 6

We get the same value of s

133 * what gets you to 432 1/4

Answers

Answer:

3 1/4

Step-by-step explanation:

You can do a equation:

133x=432 1/4

Which would give you 3.25 but in fraction form it is 3 1/4

Hope this helped!

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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