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SpyIntel [72]
3 years ago
14

Cual es el valor de 5´3

Mathematics
1 answer:
castortr0y [4]3 years ago
5 0

Answer:

I'm 5,5................

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How to graph 5x+3y=15​
maxonik [38]

Answer:

Step-by-step explanation:

The number 2 goes into both 12 and 14.

Step-by-step explanation:

12 ÷ 2 is 6 and 14 ÷2 is 7 . You can't really simplify 14 by any other whole number besides 2 or 7 .

6 0
3 years ago
4x+5y-5z=196 -11x-7y+12z=-211 -8x-9y-z=-236 Solve for x y z
FrozenT [24]
Have you learned matrices yet? I'm going to use that to solve these, please refer to the photos.

I solved the system by using a matrix calculation. The work always needs to be shown and you do that by setting up the matrix like in the first photo and also writing out all your equations. In a TI calculator, to do this you press 2ND-X^-1,GO TO EDIT - [A] - 3×4- now enter the coefficient of the system of equations variables. NOW 2ND MODE- 2ND-X^-1 -GO TO MATH- ALPHA APPS

Now, if you don't know this you may be confuse so after rref([A]) was enter a matrix came up with [1 0 0 -21] in the top row this means x equal -21. So, the next row is y and it came out as [0 1 0 46] so y equals 46 and I'm going to let you figure out what z is by looking at the matrix.

SO... X=-21 Y=46 and Z=-10

Let's Test it

4x + 5y - 5z = 196 \\ 4( - 21) + 5(46) - 5( - 10) = 196 \\  - 84 + 230 + 50 = 196 \\ 196 = 196



3 0
3 years ago
Read 2 more answers
What is the slope of line b?
Semmy [17]

Answer:

It is c)1/2

Step-by-step explanation:

Since its rising its postive

And the rise is 1 and the run is 2 so the slope is 1/2

3 0
3 years ago
Read 2 more answers
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
at the bus station, buses depart at a rate of 3 every 10 minutes, at this rate, blank buses would depart in one hour.
Musya8 [376]
18 buses would depart in an hour. 

Hope this helps :)
4 0
4 years ago
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