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Olegator [25]
3 years ago
5

Slimplify 14^0+2^7+15^1=

Mathematics
1 answer:
Leni [432]3 years ago
5 0

Answer:

1+128+15

Step-by-step explanation:

14^0 = 1

2^7 = 128

15^1 = 1

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Lesson 2 extra practice slope
lyudmila [28]
Where? I’m sorry but I can’t help if there isn’t any photo or equation/word-problem.
7 0
4 years ago
It's time for another financial calculator problem. A UCF student (who has not taken FIN 2100) decides that he really needs a la
AysviL [449]

Answer:

interest rate is 38.68 %

Step-by-step explanation:

Given data

installment = $60

time = 36 months = 36/12 = 3 years

principal = $1000

to find out

interest rate

Solution

we know student pay $60 for 36 months

so he pay total = 60 × 36 = 2160

total amount pay by student = $ 2160

so we can find interest rate by given formula

rate = (1/time)(amount/Principal - 1)

put the value time amount and principal here

rate = (1/3)(2160/1000 - 1)

rate = 0.386667

interest rate is 38.68 %

5 0
3 years ago
The fastest man in the world can run 100 metres in 9 seconds how far can He run in 1 minute and 2 seconds
finlep [7]

Answer:

Step-by-step explanation:

Rate per sec = 100 ÷ 9 = 11.111

1 min and 2 sec = 3600 + 2sec = 3602 sec

Metre = 324.21 metre

7 0
3 years ago
Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -
sattari [20]

We're looking for a solution of the form

y=\displaystyle\sum_{n\ge0}a_nx^n

with derivatives

y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n

y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n

Substituting these into the ODE gives

\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0

Shifting indices to get each term in the summand to start at the same power of x and pulling the first few terms of the resulting shifted series as needed gives

2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0

Then the coefficients in the series solution are given according to the recurrence

\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of y(0)=y'(0)=1, a degree-8 approximation of

y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).

5 0
3 years ago
Please help find the gradient :)
joja [24]

Answer:

gradient m = - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the gradient and c the y- intercept )

y = - x - 3 ← is in slope- intercept form

with gradient m = - 1

3 0
3 years ago
Read 2 more answers
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