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netineya [11]
2 years ago
14

How many terms of the arithmetic series 3+9+15.... will add up to 19200?​

Mathematics
1 answer:
Olenka [21]2 years ago
4 0

Answer:

32 005

Step-by-step explanation:

use this formula

Tn = a +(n-1)d

You might be interested in
Use graph to answer question
slamgirl [31]

Answer:

cos(θ) = 3/5

Step-by-step explanation:

We can think of this situation as a triangle rectangle (you can see it in the image below).

Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.

Here we want to find cos(θ).

You should remember:

cos(θ) = (adjacent cathetus)/(hypotenuse)

We already know that the adjacent cathetus is equal to 3.

And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:

3^2 + 4^2 = H^2

We can solve this for H, to get:

H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5

The hypotenuse is 5 units long.

Then we have:

cos(θ) = (adjacent cathetus)/(hypotenuse)

cos(θ) = 3/5

5 0
3 years ago
Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.
tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: \frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3} square units

(22)

Similar to (21)

Area = 216\sqrt{3} square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

altitude=\frac{\sqrt{3}}{2}*side

side = \frac{36}{\sqrt{3}}

Hence, area of the hexagon will be: 648\sqrt{3} square units

(24)

Given is the inradius of an equilateral triangle.

Inradius = \frac{\sqrt{3}}{6}*side

Substituting the value of inradius and calculating the length of the side of the equilateral triangle:

Side = 16 units

Area of equilateral triangle = \frac{\sqrt{3}}{4}*(side)^{2} = \frac{\sqrt{3}}{4}*256 = 64\sqrt{3} square units

4 0
3 years ago
Point J is on line segment IK. Given JK = 8 and IJ = 7, determine the length IK.​
masha68 [24]

Answer:

15

Step-by-step explanation:

If the line segment is IK, that means IJ+JK=IK. So, 7+8=15

3 0
3 years ago
Twice the complement of angles A is 40 less than the supplement of angle A. Find the measure of angle A
Serggg [28]
Complement adds to 90
complement of A is 90-A


supplement is adds to 180
supplemetn of A is 180-A

2(90-A) is -40+180-A

2(90-A)=-40+180-A
2(90-A)=140-A
distribute
180-2A=140-A
add 2A to both sides
180=140+A
minus 140 both sides
40=A

A=40 degres
4 0
3 years ago
Read 2 more answers
Could you help please..
Nitella [24]
Since she needs 3 1/2 for 5, and you need to find one batch, get 3 1/2 and divide it by 5. Change 3 1/2 to an improper fraction (7/2). Now use KCS (keep change switch) Keep the 7/2, change division to multiplication, and switch 5/1 to 1/5. (7/2)•(1/5)
4 0
3 years ago
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