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OleMash [197]
2 years ago
13

the ratio of the fourth to the first term of a geometric sequence is ⅛. if the first term exceeds the second term by 5, find the

first and the 8th term of the sequence?​
Mathematics
1 answer:
____ [38]2 years ago
5 0

Answer:

a = 10

T8 = 1280

Step-by-step explanation:

The nth term of a GP is expressed as;

Tn = ar^n-1

Forth term T4 = ar^3

first term = a

If the ratio of the fourth to the first term of a geometric sequence is ⅛ then;

ar^3/a = 1/8

r^3 = 1/8

r = ∛1/8

r = 1/2

If the first term exceed the second by 5, then;

a = 5 + T2

a = 5 + ar

a = 5 + a(1/2)

a-1/2a = 5

1/2 a  5

a = 10

Hence the first term is 10

T8 = ar^7

T10 = 10(1/2)^7

T10 = 10(128)

T10 = 1280

Hence the 8th term is 1280

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Estimate the perimeter and the area of the shaded figure to the nearest whole number.
soldier1979 [14.2K]

Answer:

Step-by-step explanation:

The area and perimeter of the given shaded figure are respectively 33.12 unit² and 20.56 units.

Area and perimeter-based problem:

What information do we have?

Radius of semi-circle = 8 / 2 = 4 unit

Length of remain Rectangle = 8 unit

Width of remain Rectangle = 2 unit

Perimeter of shaded figure = πr + (l + 2b)

Perimeter of shaded figure = (3.14)(4) + [4 + (2)(2)]

Perimeter of shaded figure = 12.56 + 4 + 4

Perimeter of shaded figure = 20.56 units

Area of shaded figure = πr²/2 + lb/2

Area of shaded figure = (3.14)(4)²/2 + (8)(2)/2

Area of shaded figure = (3.14)(8) + 8

Area of shaded figure = 25.12 + 8

Area of shaded figure = 33.12 unit²

5 0
2 years ago
Read 2 more answers
Study the figure.
tia_tia [17]

Answer:

So we know the formula to calculate the area of the circular sector:

S=(r^2*π*a)/306°=

(5^2*3.14*40°)/360°= (1000*3.14) /360=8.72cm^2 so the right alternative should be the first one ,A.

Step-by-step explanation:

r - radius of the circle

a - corner of the circular sector

S - surface

4 0
3 years ago
Find the value of x - if x^3-7x² - x = 0<br>a) 17 (b) 71 c) 7<br>d) 1​
S_A_V [24]

Answer:

x = 0

Step-by-step explanation:

x^3-7x^2-x =0\\

Take out common factor

x(x^2-7x-1) = 0

The factor inside the parenthesis cannot be factored

Since there is a common factor that we pulled out of the original equation,

set that x equal to zero and solve

x = 0\\ : Therefore x = 0

6 0
3 years ago
What are the zeos of f(x)=(x-1)(x+6)
solong [7]

The zeroes of the given polynomial is 1 and -6.

<h3>What is Polynomials?</h3>

A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Here, given equation;

          f(x) = (x - 1)(x + 6)

for finding the zeros of the equation we need to put.....f(x) = 0

           0 = (x - 1)(x + 6)

     x - 1 = 0      or       x + 6 = 0

     x = 1          or    x = -6

Thus, The zeroes of the given polynomial is 1 and -6.

Learn more about Polynomials from:

brainly.com/question/17822016

#SPJ1

7 0
2 years ago
Exhibit B: A restaurant has tracked the number of meals served at lunch over the last four weeks. The data shows little in terms
Mekhanik [1.2K]

Answer:

Option E is correct.

The expected number of meals expected to be served on Wednesday in week 5 = 74.2

Step-by-step Explanation:

We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.

Week

Day 1 2 3 4 | Total

Sunday 40 35 39 43 | 157

Monday 54 55 51 59 | 219

Tuesday 61 60 65 64 | 250

Wednesday 72 77 78 69 | 296

Thursday 89 80 81 79 | 329

Friday 91 90 99 95 | 375

Saturday 80 82 81 83 | 326

Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952

Total number of meals served at lunch on Wednesdays over the 4 weeks = 296

Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443

Total number of meals expected to be served in week 5 = 490

Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3

Checking the options,

74.3 ≈ 74.2

Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2

Hope this Helps!!!

8 0
2 years ago
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