Ask question

Ask question

All categories

3 years ago

6

What is the nth ternm of this sequence -4,1,6,11,16, a 5n-4 b -4n+5 c -9n d 5n-9

2 answers:

vfiekz [6]3 years ago

8 0

Answer:

d. 5n-9

Step-by-step explanation:

<u>Given sequence:</u>

-4, 1, 6, 11, 16, ...

<u>It is an AP with:</u>

The first term is a = -4
Common difference is d = 5

<u>Nth term of this sequence is:</u>

aₙ = a + (n - 1)d
aₙ = - 4 + 5(n - 1) = -4 + 5n - 5 = 5n - 9

Correct option is d.

Nookie1986 [14]3 years ago

7 0

Answer:

<h2>d. (5n-9)</h2>

Step-by-step explanation:

First term = -4
Common difference = 1-(-4)=1+4=5
Formula used: L =a+(n-1)d

where, L is the last term;

a is the first term ;

n is the number of terms;

and d is the common difference.

<em>L = -4+(n-1)5= -4+5n-5 = 5n-9</em>

Therefore, the last term is (5n-9).

You might be interested in

on Wednesday jean's nursery received a shipment of 60 flowering crabapple trees. jean has ordered 80 trees. what percent of her

WITCHER [35]

75% when you have 60/80

7 0

4 years ago

Which of the following numbers could be the sides of a right triangle. A)8,15,17. B)6,8,12 C)9,11,21 D)4,13,16

ArbitrLikvidat [17]

Answer:

A) 8, 15, 17.

Step-by-step explanation:

Right triangle obey the Pythagorean theorem. Thus, we choose the two smaller numbers (being the cathetus) and if after applying the P. Theorem we get the biggest of each option (the hypotenuse) that means that those numbers could be the sides of a right triangle.

The Pythagorean theorem states that: $a^2+b^2=c^2$

Thus:

$\sqrt{a^2+b^2}=c$

Option A:

$\sqrt{8^2+15^2}=17$ → 17 = 17 OK!

Option B:

$\sqrt{6^2+8^2} = 10$ → 10 ≠ 12 NO

Option C:

$\sqrt{9^2+11^2} = [tex]\sqrt{202}$ [/tex] → $\sqrt{202}$ ≠ 21 NO

Option D:

$\sqrt{4^2+13^2} = \sqrt{185}$ → $\sqrt{185}$ ≠ 16 NO

3 0

3 years ago

Pls help need help ASAP 50 points you will be reported if the answer is wrong

aev [14]

Answer:

Part A = 10in = 35 ft

= 20 = 70 ft

Part B =

$ 178.5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Use the given minimum and maximum data entries, and the number of classes, to find the class width, the lower class limits,

Gennadij [26K]

Using proportions and the information given, it is found that:

The class width is of 14.375.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

Minimum value is 19.
Maximum value is of 134.
There are 8 classes.
The classes are all of equal width, thus the width is of:

$W = \frac{134 - 19}{8} = 14.375$

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

6 0

3 years ago

Plllzzz im new and i neeed help

ella [17]

Answer:

4082

Step-by-step explanation:

Given

The composite object

Required

The volume

The object is a mix of a cone and a hemisphere

Such that:

<u>Cone</u>

$r = 10cm$ ---- radius (r = 20/2)

$h = 19cm$

<u>Hemisphere</u>

$r=10cm$

The volume of the cone is:

$V_1 = \frac{1}{3}\pi r^2h$

$V_1 = \frac{1}{3}\pi * 10^2 * 19$

$V_1 = \frac{1900}{3}\pi$

The volume of the hemisphere is:

$V_2 = \frac{2}{3}\pi r^3$

$V_2 = \frac{2}{3}\pi 10^3$

$V_2 = \frac{2000}{3}\pi$

So, the volume of the object is:

$V = V_1 + V_2$

$V = \frac{1900}{3}\pi + \frac{2000}{3}\pi$

$V = \frac{3900}{3}\pi$

$V = 1300\pi$

$V = 1300 * 3.14$

$V = 4082$

8 0

3 years ago