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3 years ago

At which point do the two equations

Mathematics

1 answer:

Lena [83]3 years ago

4 0

The lines intersect at two places.

They intersect at approximately (-2,8) and (2,3)

Looking at the choices they would be at (1.8,3.2) and (-2.8,7.8)

The answer is D. Both A and B.

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Question is below help needed

RoseWind [281]

The common ratio of both the formulas is 3 and 2. The first formula is explicit and the second is the recursive formula.

<h3>What is the recursive rule?</h3>

A rule is defined such that its definition includes itself.

The given sequence is

-2, -6, -18, -54

Common ratio ;

r = -6 / -2 = 3

$a _ n =-2 (3) ^{n-1}$

This given formula is the explicit formula.

The given sequence is

4, 8, 16, 32

Common ratio ;

r = 8 / 4 = 2

$a_n = a_{n-1}.2\\\\a _1 = 4$

This given formula is the recursive formula.

Learn more about the recursive rule and explicit rules here:

brainly.com/question/12620593

#SPJ1

5 0

2 years ago

What is the area of 4.5 and 5.5cm and 20cm

amm1812

Answer:

6.8

Step-by-step explanation:

8 0

3 years ago

A triangle is shown what is the length in inches of side a

amm1812

I think it’s 74 bc I got it right but your might be different

4 0

3 years ago

PLEASE HELP!! I WILL BRAINLIEST

ArbitrLikvidat [17]

Answer:

81.8%

Step-by-step explanation:

Mean = $\mu = 40$

Standard deviation = $\sigma = 5$

Now we are supposed to find out what percent of the numbers fall between 35 and 50

$z = \frac{x-\mu}{\sigma}$

Substitute the values

$z = \frac{x-40}{5}$

Now for P(35<x<50)

Substitute x = 35

$z = \frac{35-40}{5}$

$z =-1$

Substitute x = 50

$z = \frac{50-40}{5}$

$z =2$

So, P(-1<z<2)

P(z<2)-P(z<-1)

=0.9772-0.1587

=0.8185

= $0.818 \times 100$

=81.8%

Hence 81.8% percent of the numbers fall between 35 and 50

7 0

3 years ago

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

3 years ago