MARKING BRAINLIST IF THESE TWO ARE ANSWERED‼️‼️‼️

Does the following infinite series converge or diverge? 1/3+2/9+4/27+8/81+...

PilotLPTM [1.2K]

Note that A and D are ludicrous choices, so you can throw them away outright. (Any divergent series cannot have a sum, and any convergent series must have a sum.)

The sum is certainly convergent because it can be written as a geometric sum with common ratio between terms that is less than 1 in absolute value.

$S=\dfrac13+\dfrac29+\dfrac4{27}+\dfrac8{81}+\cdots$
$S=\dfrac13\left(1+\dfrac23+\dfrac{2^2}{3^2}+\dfrac{2^3}{3^3}+\cdots\right)$

We can then find the exact value of the sum:

$\dfrac23S=\dfrac13\left(\dfrac23+\dfrac{2^2}{3^2}+\dfrac{2^3}{3^3}+\dfrac{2^4}{3^4}+\cdots\right)$

$\impliesS-\dfrac23S=\dfrac13$
$\implies\dfrac13S=\dfrac13$
$\implies S=1$

So the answer is B.

8 0

3 years ago

Which number line shows the approximate location of the square root of 10?

Aleks [24]

Answer:

It should be somewhere near 3.16

Step-by-step explanation:

The square root of 10 simplified is around 3.16

5 0

3 years ago

Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a

evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1= $\frac{1}{2} *PQ*RM$

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2= $\frac{1}{2} *SR*PN$

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1= $\frac{1}{2}*PQ*h$

S2= $\frac{1}{2}*SR*h$

Adding them together, we get the area of the whole trapezoid:

S=S1+S2= $\frac{1}{2} (PQ+SR)h,$

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> $A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}$

Simplifying, we have

=> $A= \frac{ah+2b_{1+C} }{2}$

Factoring we have,

=> $A_{PQRS} = (a+ 2b_{1} + c)\frac{h}{2} \\= > {(a+ b_{1} + c) + b_{1} }\frac{h}{2}$

But, $a+ b_{1} + c$ is equal to $b_{2}$ , the longer base of our trapezoid.

Hence, $A_{PQRS}= (b_{1} + b_{2} )\frac{h}{2}$

We have discussed two ways by which we can derive area of a trapezoid.

Read to know more about Trapezoid

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5 0

2 years ago

Plss help me TT

Marina CMI [18]

a. There are 0 or 2 real positive roots for the equation and

b. There are 0 or 2 real negative roots for the equation.

<h3>What is the Descartes'rule of sign?</h3>

Descartes' rule of sign states that

The number of real positive zero of a polynomial f(x) is the number of sign changes of the coefficients of f(x) or an even number less than the number of sign changes of the coefficients of f(x)
The number of real negative zero of a polynomial f(x) is the number of sign changes of the coefficients of f(-x) or an even number less than the number of sign changes of the coefficients of f(-x)

<h3>How to find the number of possible positive and negative roots are there for the equation?</h3>

Given the equation 0 = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1, writing it as a polynomial function, we have f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1

<h3>a. The number of positive roots</h3>

So, to find the number of positive roots, we find the number of sign changes of the polynomial f(x).

So, f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1

Since f(x) has coefficients -8, -2, + 8, -4, -1, there are two sign changes from -2 to + 8 and from + 8 to -4.

So, there are 2 or 2 - 2 = 0 real positive roots.

So, there are 0 or 2 real positive roots for the equation.

<h3>b. The number of negative roots</h3>

So, to find the number of negative roots, we find the number of sign changes of the polynomial f(-x).

So, f(-x) = −8(-x)¹⁰ − 2(-x)⁷ + 8(-x)⁴ − 4(-x)² − 1

= −8x¹⁰ + 2x⁷ + 8x⁴ − 4x² − 1

Since f(x) has coefficients -8, +2, + 8, -4, -1, there are two sign changes from -8 to + 2 and from + 8 to -4.

So, there are 2 or 2 - 2 = 0 real negative roots.

So, there are 0 or 2 real negative roots for the equation.

So,

There are 0 or 2 real positive roots for the equation and
There are 0 or 2 real negative roots for the equation.

Learn more about Descartes' rule of sign here:

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8 0

2 years ago

Please answer 6! Thank you!

Jobisdone [24]

Answer:

∠J= 22°

JL=19.0137

KL=7.99

Step-by-step explanation:

The step-by-step is attached as a picture. Hope it helps! Feel free to message me if you have any questions in reference to this math problem.

The answer to the problem you sent me in the comments is attached to here too.

4 0

3 years ago