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Digiron [165]
2 years ago
6

Pls help I will give you 15 points only

Mathematics
1 answer:
QveST [7]2 years ago
6 0
N= 1/8 I’m pretty sure I did that last year
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Solve the system of equations. -10y+9x=-9 <br> 10y+5x=-5
alexira [117]

by the use of elimination method

make all coefficients of subject to be eliminated similar..by multiplying the coefficients with one another

for eqn(i)

5(-10y+9x=-9)

-50y+45x=-45

for eqn(ii)

9(10y+5x=-5)

90y+45x=-45

-50y+45x=-45

90y+45x=-45

...subtract each set from the other...

we get

-140y+0=0

y=0

from eqn(i)

10y+5x=-5

0+5x=-5

x= -1

3 0
3 years ago
Solve 3[-x + (2 x + 1)] = x - 1.
Marrrta [24]

Answer:

x= -2

Step-by-step explanation:

-3x+6x+3=x-1

3x-x= -1-3

2x= -4

x= -2

6 0
3 years ago
Read 2 more answers
Which of these sequences is a geometric sequence?
Harman [31]

Answer:

d is right...............

5 0
3 years ago
Write p(x) = 21 + 24x + 6x2 in vertex form.
Helen [10]
To do this, complete the square:

p(x) = 21 + 24x + 6x2  =>  <span>p(x) = 6x2 + 24x + 21

Rewrite the first 2 terms as


                                                   6(x^2 + 4x)

then you have </span><span>p(x) =  6(x2 + 4x                           ) + 21
                            
Now complete the square of x^2 + 4x:

                         p(x) = 6(x^2 + 4x + 4 - 4)            + 21
                                 = 6(x+2)^2 - 24 + 21

                          p(x)  = 6(x+2)^2 - 3        this is in vertex form now.

We can read off the coordinates of the vertex from this:  (-2, -3)</span>
8 0
3 years ago
Find the radius of convergence, then determine the interval of convergence
galben [10]

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = β

  • If β < 1, then the series converges
  • If β > 1, then the series diverges
  • If β = 1, then the series inconclusive

Here a_{k} = \frac{(x+2)^{k}}{\sqrt{k} }  and  a_{k+1} = \frac{(x+2)^{k+1}}{\sqrt{k+1} }

   

Now limit is taken,

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|

= \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|

= \lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|

= |{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}

= |{x+2 }| < 1

- 1 < {x+2 } < 1

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

 

We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

brainly.com/question/14394994

#SPJ1

5 0
1 year ago
Read 2 more answers
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