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harkovskaia [24]
3 years ago
12

Two fractions negative whose product is 5/8

Mathematics
1 answer:
Andreyy893 years ago
8 0
You could do -1/-2 times -5/-4 which would be 5/8
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10. A local
professor190 [17]

Answer:

39.7%

Step-by-step explanation:

(actual value - estimated value) / actual value

5.8 - 3.5 = 2.3

2.3 / 5.8 = 0.3965 = 39.65%

5 0
3 years ago
Do sb know da answerrrrr ?
shutvik [7]

Answer:

If I’m not wrong I think it’s the first one

Step-by-step explanation:

8 0
3 years ago
Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z​
Oliga [24]

<u>Part a)</u>

Given the expression

54m^3n\:+\:81m^4n^2

Apply exponent rule:    a^{b+c}=a^ba^c

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn     ∵ m^4n^2=m^3mnn

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

                             =2\cdot \:27m^3n+3\cdot \:27m^3mnn

Factor out the common term:   27m³n

                              =27m^3n\left(2+3mn\right)      

Therefore,

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)

<u>Part B)</u>

Given the expression

15x^2y^3z+25x^3y^2z+35x^2y^2z

Apply exponent rule:    a^{b+c}=a^ba^c

15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z

Rewrite as

                                               =3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z

Factor out common term 5y²x²z

                                                =5y^2x^2z\left(3y+5x+7\right)

Therefore,

15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)

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
What is the quotient?<br><br> x + 1)3x2 − 2x + 7
melisa1 [442]

Answer:

3x-5

Step-by-step explanation:

8 0
3 years ago
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