All categories

3 years ago

What is the explicit rule for the geometric sequence? 4.05,1.35,0.45,0.15,...

Mathematics

1 answer:

Anastaziya [24]3 years ago

4 0

Step-by-step explanation:

Since the above sequence is a geometric sequence

For an nth term in a geometric sequence

That's

$A(n) = a( {r})^{n - 1}$

where

r is the common ratio

a is the first term

n is the number of terms

From the question

a = 4.05

r = 1.35 / 4.05 = 0.33

Substitute the values into the above formula

We have

$A(n) = 4.05( {0.33})^{n - 1}$

Hope this helps you

You might be interested in

Please help me Due in an hour

vladimir1956 [14]

131.88 or jus do radius times 2 so 21+ 21 = 42.

42 x pi(3.14) = 131.88

3 0

3 years ago

A square piece of paper is folded in half vertically. The resulting figure has a perimeter of 21 in. What is the number of units

Rudiy27

Perimeter is adding all 4 sides together.

To find the length and width divide total perimeter by 2:

21 / 2 = 10.5

We now know the length of 1 side and the width of 1 side need to equal 10.5.

Since the original piece was a square we also know the width needs to equal 1/2 of the length.

So we have L + 1/2L = 10.5

Add the common terms 1 L + 1/2 L = 1 and 1/2 L

To Find L divide 10.5 by 1-1/2, which equals 7

To find the width subtract 7 from 10.5, which equals 3.5

So the paper is now 3.5 inches by 7 inches.

4 0

3 years ago

Can someone please explain how to do this!!!

DedPeter [7]

Hi there!

So we are given that:-

tan theta = 7/24 and is on the third Quadrant.

In the third Quadrant or Quadrant III, sine and cosine both are negative, which makes tangent positive.

Since we want to find the value of cos theta. cos must be less than 0 or in negative.

To find cos theta, we can either use the trigonometric identity or Pythagorean Theorem. Here, I will demonstrate two ways to find cos.

Using the Identity

$\large \displaystyle{ {tan}^{2} \theta + 1 = {sec}^{2} \theta}$

Substitute tan theta = 7/24 in.

$\large \displaystyle{ {( \frac{7}{24}) }^{2} + 1 = {sec}^{2} \theta }$

Evaluate.

$\large \displaystyle{ \frac{49}{576} + 1 = {sec}^{2} \theta} \\ \large \displaystyle{ \frac{625}{576} = {sec}^{2} \theta}$

Reminder -:

$\large \displaystyle{ sec \theta = \frac{1}{cos \theta} }$

Hence,

$\large \displaystyle{ \frac{576}{625} = {cos}^{2} \theta} \\ \large \displaystyle{ \sqrt{ \frac{576}{625} } = cos \theta} \\ \large \displaystyle{ \frac{24}{25} = cos \theta}$

Because in QIII, cos<0. Hence,

$\large \displaystyle \boxed{ \blue{cos \theta = - \frac{24}{25} }}$

Using the Pythagorean Theorem

$\large \displaystyle{ {a}^{2} + {b}^{2} = {c}^{2} }$

Define c as our hypotenuse while a or b can be adjacent or opposite.

Because tan theta = opposite/adjacent. Therefore:-

$\large \displaystyle{ {7}^{2} + {24}^{2} = {c}^{2} } \\ \large \displaystyle{49 + 576 = {c}^{2} } \\ \large \displaystyle{625 = {c}^{2} } \\ \large \displaystyle{25 = c}$