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4 years ago

What is the value of the function y=3x−1 when x=−1 ? −8 −4 2 6

Mathematics

2 answers:

lara31 [8.8K]4 years ago

6 0

Answer:

Option B is correct

Value of the function $y =3x-1$ when x =-1 is -4

Step-by-step explanation:

Given the function: $y =3x-1$ ......[1]

Here, x is the input value and y is the output value.

Substitute the input value of x = -1 in [1]

$y =3(-1)-1$

⇒ $y =-3-1$

⇒ $y =-4$

Therefore, the value of the function $y =3x-1$ when x = -1 is -4

Paraphin [41]4 years ago

3 0

Substitute x for -1 in y=3x-1
y = (3) (-1) -1
Solve
y = (3) (-1) -1
y = -3 - 1
y = -4

B.) -4 is your answer

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The additive inverse of -40

Genrish500 [490]

Answer:

Step-by-step explanation:

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3 years ago

PLEASE HELP ME ASAP!!!!!!! TY! :)

tekilochka [14]

Answer:

2 nickels and 21 dimes

Step-by-step explanation:

0.05N + 0.1D = 2.2

N + D = 23

Let's use the substitution method. Solve the second equation for N.

N = 23 - D

Now where you see N in the first equation substitute N with 23 - D.

0.05(23 - D) + 0.1D = 2.2

1.15 - 0.05D + 0.1D = 2.2

0.05D = 1.05

D = 1.05/0.05

D = 21

Now substitute 21 for D in the second original equation and solve for N.

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3 years ago

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amm1812

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you then count every number after 3 including 8 which is

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3 years ago

(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also c

valina [46]

Answer:

a) See the proof below.

b) $\sum \frac{(-x)^n}{n}$

Step-by-step explanation:

Part a

For this case we assume that we have the following series $\sum a)n x^n$ and this series has a finite radius of convergence $R$ and we assume that $a_n \geq 0$ for all n, this information is given by the problem.

We assume that the series converges at the point $x= R$ since w eknwo that converges, and since converges we can conclude that:

$\sum a)n R^n < \infty$

For this case we need to show that converges also for $x=-R$

So we need to proof that $\sum a_n (-R)^n < \infty$

We can do some algebra and we can rewrite the following expression like this:

$\sum a_n (-R)^n = \sum (-1)^n a)n R^n$ and we see that the last series is alternating.

Since we know that $\sum a_n x^n$ converges then the sequence { $a_n R^n$ } must be positive and we need to have $lim_{n\to \infty} a^n R^n = 0$

And then by the alternating series test we can conclude that $\sum a_n (-R)^n$ also converges. And then we conclude that the power series $a_n x^n$ converges for $x=-R$ ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this $\frac{(-x)^n}{n}$

Because the series $\sum \frac{(-x)^n}{n}$ converges to $-ln(1+x)$ when $|x|$ using the root test.

But by the properties of the natural log the series diverges at $x=-1$ because $\sum \frac{1}{n} =\infty$ and for $x=1$ we know that converges since $\sum \frac{-1}{n}$ is an alternating series that converges because the expression tends to 0.

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