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2 years ago

What common ratio would cause a geometric series to converge

Mathematics

1 answer:

iVinArrow [24]2 years ago

6 0

Answer:0

Step-by-step explanation:

The sum of a convergent geometric series can be calculated with the formula a ⁄ 1 – r, where “a” is the first term in the series and “r” is the number getting raised to a power. A geometric series converges if the r-value (i.e. the number getting raised to a power) is between -1 and 1. So the answer is 0

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Change the polar coordinates to rectangular coordinates. (a) (3, 5π/6)

elixir [45]

Ok so given the point (r, theta)
The corresponding Cartesian point is (r*sin(theta), r* cos(theta)) you can think about this by analyzing the points on a unit circle which is a graph of a polar circle with radius 1 and angle theta

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3 years ago

Someone find area and perimeter of this shape! Thanjs

Gemiola [76]

The area would be 132cm^2 (squared)
(9x4)x2+(7+8)x4 =132

The perimeter you just add everything up so it’s 65cm
15+13+4+7+9+4+13

6 0

3 years ago

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Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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2 years ago

The formula for the volume of a cone is V = jarh, where V is the volume, r is the radius, and h Is the heightWhat is the volume

olga nikolaevna [1]

Given:

The radius, r=4x

The height,

$h=21xy^2$

To find the volume of the cone:

The volume of the cone formula is,

$V=\frac{1}{3}\pi r^2h$

Substitute the values of r and h in the above formula we get,

$\begin{gathered} V=\frac{1}{3}\pi\times(4x)^2\times21xy^2 \\ =\frac{1}{3}\pi\times16x^2\times21xy^2 \\ =\pi\times16x^2\times7xy^2 \\ =112\pi x^3y^2 \end{gathered}$

Hence, the volume of the cone is

$V=112\pi x^3y^2$

Thus, the correct option is option D.

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8 months ago

What is the missing justification in the proof?

nataly862011 [7]

The missing justification in the proof is
<span>B) Substitution property of equality

The expression for sin</span>² x and cos² x is substituted to the other side of the equation. Since sin x = a/c, then sin² x = a²/c². Similarly, since cos x = b/c, then cos² x = b²/c². Adding to two results to the third statement.

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3 years ago

Read 2 more answers