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3 years ago

In arithmetic sequence the first term is 5 and the common difference is 4. Determine the 6th term of sequence

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

3 0

Answer:

The 6th term of the sequence is 25

Step-by-step explanation:

first term = a = 5

common difference = d = 4

The nth term of a sequence is

T_n = a + (n - 1)d

Therefore, the 6th term of the sequence is

T_6 = a + (6 - 1)d

T_6 = a + 5d

T_6 = 5 + 5(4)

T_6 = 5 + 20

T_6 = 25

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20 POINTS FOR RIGHT ANSWER!!!

dangina [55]

Answer:

a=10 and b= -12

Step-by-step explanation:

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3 years ago

At a high school the student government is made up of 5 seniors, 4 juniors, 3 sophomores and 2 freshmen. If there are 8 seniors,

hjlf

Answer:

32598720

Step-by-step explanation:

<u>Combination for the school government requires:</u>

5 out of 8 seniors = 8! / 3!5! = 6*7*8/2*3 = 7*8 = 56
4 out of 9 juniors = 9! / 4!5! = 6*7*8*9/2*3*4= 7*2*9 = 126
3 out of 12 sophomores = 12! / 3!9! = 10*11*12/2*3 = 10*11*2 = 220
2 out of 7 fishermen = 7! / 2!5! = 6*7/2 = 21

<u>So number of total ways is:</u>

56*126*220*21 = 32598720

5 0

2 years ago

A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

$m\angle AOM =30^o$

In the right triangle OAM

$tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}$

$\frac{a}{2r}=\frac{\sqrt{3}}{3}$

we have

$r=3\ units$

substitute

$\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units$

Find the area of six equilateral triangles

$A=6[\frac{1}{2}(r)(a)]$

simplify

$A=3(r)(a)$

we have

$r=3\ units\\a=2\sqrt{3}\ units$

substitute

$A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2$

Therefore

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

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3 years ago

Gia bought a computer that was 20% off the regular price of $1,280. If a 9% sales tax was added to the cost of the computer, wha

garik1379 [7]

1280*.8=1024
1024*1.09=1116.16
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3 years ago

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Which is the equation of a line with a slope of −13 - 1 3 and a y-intercept at (0, 1) 0 , 1 ?

Natasha_Volkova [10]

Rawr . Rawr . Rawr .

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2 years ago