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Alex777 [14]
2 years ago
11

Can you guys help out I these questions for a test

Mathematics
1 answer:
DaniilM [7]2 years ago
5 0

the first x < 3 so x could be from - 1 till 3 in this graph

the other x > - 7 so x could be from from - 7 till - 3 in this graph

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Simplify: 3(6+6i)+i(6+6i)
GuDViN [60]

Answer:

the answer is 12+24i            

                               

6 0
3 years ago
What is the area of ABC below? If necessary, round your answer to two decimal places.
irina [24]
Our goal is to find the area
area=1/2 times base times height
we know the height is 7.9 but what is the base?


use the pythagorean theorem twice
alrighty
so
remember for legs length a and b and hytponuse c in a right triangle
a²+b²=c²

we need AD and DC

so
AD²+7.9²=9.4²
AD=√25.95

DC²+7.9²=23.2²
DC=√475.83

so
AD+DC=base=(√25.95)+(√475.83)
area=1/2bh
area=1/2((√25.95)+(√475.83))(7.9)
area≈106.28518654591426812803776879893
round to 2 decimal places
area≈106.29 square units
6 0
2 years ago
The sum of the interior angles of a regular pentangon
natima [27]

Answer:

540

Step-by-step explanation:

ARCHITECTURE The Pentagon in Washington, D.C. is shaped like a regular pentagon. Find the sum of the measures of the interior angles of the largest pentagon-shaped section of the Pentagon building. = 180(3) or 540 Simplify. The sum of the measures of the interior angles is 540.

4 0
2 years ago
Read 2 more answers
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
3 years ago
Can someone pls help :(
aleksandrvk [35]

Answer:

srry man i suck at civics, but i think it is B?

Step-by-step explanation:

7 0
3 years ago
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