MATHEMATICS COLLEGE

Help me pls thank you

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

To solve this, we have to solve for h.

We can use order of operations to figure this one out.

$18h-9h-7h+3=17\n18h-9h-7h=14$

Now that every variable on the left side of the equation has an h, we can factor.

$h(18-9-7)=14\nh=(14)/(18-9-7) \nh=(14)/(2) \nh=7$

Answer 2

Answer:

Answer:

I just did some trial and error with the numbers

186-96-76+3=17

Unless I did it wrong, then just ignore this answer

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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

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

How much change
would you get from £1
if you spent 77p?

Answers

You would get 23p as change from £1 if you spent 77p.

To find how much change you would get from £1 if you spent 77p, you need to subtract the amount you spent from the total amount you have.

Total amount = £1

Amount spent = 77p

To find the change, we need to convert both amounts to the same unit (pence) before performing the subtraction.

1 pound (£1) is equal to 100 pence, so:

Total amount = 100p

Amount spent = 77p

Now, subtract the amount spent from the total amount to find the change:

Change = Total amount - Amount spent

Change = 100p - 77p

Change = 23p

So, you would get 23p as change from £1 if you spent 77p.

To know more about change:

brainly.com/question/16545971

#SPJ2

You would get 23p change

|-9×+7|+8 is less than or equal to 9

Answers

Answer:

I don’t think it’s neither less than it equal

Step-by-step explanation:

I could be wrong don’t listen to me :)

Which expression is equivalent to b^m x b^n? A. b^m+n B. b^m-n C. b^mx n D. b^m/n

Answers

Answer:

The expression which is equivalent to

$\rightarrow b^m * b^n\n\n=b^(m+n)\n\n\rightarrow x^a*x^b=x^(a+b)$

Option A

The formula for multiplying exponents are such below.

(b^m)^n = b^mn
b^m/b^n=b^(m-n)
b^m x b^n=b^(m+n)

Could someone please help?

Find the perimeter of each figure.

Answers

Answer:

3.6

Step-by-step explanation:

stupid idiot you add them up together go back to 4th grade

Is {(98,6),(99,0),(100,7)} a function?

Answers

Answer:

Yes , this is the function

Step-by-step explanation:

Because every domain has their only one image

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