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3 years ago

Whats the first step when solving for g? 6g-4=8

2 answers:

8 0

Here's how to sole for g in steps:

6g=12------------------------------------Step 1: Add 4 to both sides

g=2---------------------------------------Step 2: Divide both sides by 6

Good luck with your classes! I hope this helped!

neonofarm [45]3 years ago

8 0

The first step would be to add 4 to both sides of the equation.

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What is the area of figure shown below!??

Novosadov [1.4K]

Answer:26

Step-by-step explanation:

Add them all up

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3 years ago

What is the answer plzzzzzzzzzzzzzzz I need help

Licemer1 [7]

Answer:

(11 + c) + 8

Step-by-step explanation:

Step 1

The question says "11 + Carlos's score (represented with a c)".

Hence: 11 + c

Step 2

The question says "eight more than 11 + Carlos's score".

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4 years ago

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1. What's 7/8 as a decimal?

denpristay [2]

Answer:

0.875

Step-by-step explanation:

7/8 as a decimal is 0.875. Another was of thinking of this is the number '7' is 87.5 percent of the number '8.

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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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3 years ago

Consider the paragraph proof. Given: D is the midpoint of AB, and E is the midpoint of AC. Prove:DE = BC It is given that D is t

denis23 [38]

It is not D and d was 4a^2

4 0

3 years ago

Read 2 more answers