All categories

2 years ago

What is the complement of 30 degrees

Mathematics

2 answers:

cestrela7 [59]2 years ago

7 0

30° is the compliment

ivolga24 [154]2 years ago

5 0

The complement of 30 degrees is 60

Why?

A complement is a thing that completes something.

Now if your angle is 30 degrees, to complete the angle to be a right angle will be 60. Because 30 + 60 will equal to 90 degrees.

Hope this helped!

You might be interested in

Pls Help, 20 points!

skelet666 [1.2K]

Answer:

90/13, 45/13

Step-by-step explanation:

$- 2 = x \\ 7x + 9y = - 45 \\ \\ 7 \times ( - 2)y + 9y = - 45 \\ y = 9 \\ \\ 7x + 9 \times 9 = - 45 \\ x = - 18 \\ (x.y) = (- 18.9) \\ \\$

Simplify:

$- 2 \times 9 = - 18 \\ 7 \times ( - 18) + 9 \times 9 = - 45 \\ - 18 = - 18 \\ - 45 = - 45$

5 0

2 years ago

Read 2 more answers

Two spheres M and N have volumes of 500 cubic cm and 9500 cubic cm respectively. Find the ratio of their radii.

frosja888 [35]

Answer:

4.9:13.1

Step-by-step explanation:

V=4/3*pi*r^3

8 0

2 years ago

PLEASE ..... HELP THANK YOU

lara [203]

5x-10=3x+40
5x=3x+50
2x=50
x=25
5*25=125
125-10=115
angle AEB=115

Hope this helps :)

4 0

2 years ago

Read 2 more answers

(girls only) drop ur ig (I’m a girl btw :)))

andriy [413]

Step-by-step explanation:

iiputriii:)

pa brainlist nga po

7 0

1 year ago

The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

4 0

2 years ago