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3 years ago

What is the value of b in the triangle shown below?

Mathematics

2 answers:

Akimi4 [234]3 years ago

7 0

Answer:

b= 4 in.

Step-by-step explanation:

Ok. So the area of a triangle is base * height divided by 2.

So, the base is b and the height is 3b.

b*3b= 3b^2.

Now, divide that by 2.

(3b^2)/2.

Now, we know what the area is already- 24 in.^2. So, let's set the area expression that I found earlier equal to this area, since they're the same.

(3b^2)/2 = 24

3b^2 = 48

b^2 = 16

Ok. Let's stop here. Now, we square root both sides to obtain b, but we are presented with a problem here. B can equal both 4 and -4 right? Since -4*-4=16 and 4*4=16. However, it is impossible for a 2d shape to have negative measurements. So, the answer is 4.

12345 [234]3 years ago

4 0

Answer:

+4 in

Step-by-step explanation:

A= 1/2 x b x h

24=1/2 x b x 3b

b2=16

b=+4 or -4(N/A)

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Step-by-step explanation:

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3 years ago

The surface area of a cube is 216in.² what is the volume of a cube

saw5 [17]

Answer:216 in^3

Step-by-step explanation:

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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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Lapatulllka [165]

Answer:

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Step-by-step explanation:

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