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Anastaziya [24]
3 years ago
8

Find the area of the following figure. Explain or show how you found your answer.

Mathematics
1 answer:
zaharov [31]3 years ago
3 0

Answer:No clue

Step-by-step explanation:

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Find the perimeter of the isosceles trapezoid
Genrish500 [490]

Answer:

A =136 cm^2

Step-by-step explanation:

The area of a trapezoid is given by

A = 1/2 (b1+b2) *h  where b1 and b2 are the lengths of the bases and h is the height

A = 1/2( 22+12) *8

A = 1/2 ( 34)*8

A =136 cm^2

4 0
2 years ago
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What is the length of AC?
Korvikt [17]
AB+BC you will get the length of AC
8 0
3 years ago
Aisle 2 of a furniture store stocks 4-legged chairs, 4-legged tables, and 3-legged stools. If there are t tables, c chairs, and
ruslelena [56]

4tx+4cx+3sx=41

You say 4x4=16

+4x4            =16

+3x3                9

5 0
3 years ago
type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar. the square of a num
lorasvet [3.4K]
Let,
The number be "x"

According to the question,

x² + 5x = 176
x² + 5x - 176 = 0
x² + 16x - 11x - 176 = 0
x (x + 16) - 11 ( x + 16) = 0
(x + 16) (x - 11) = 0

Now,
Using zero product property:

Either,
x + 16 = 0
x = -16

Or,
x - 11 = 0
x = 11

So,
The number is either "-16" or "11".


6 0
3 years ago
Please help me answer this question
strojnjashka [21]

Answer:

  • modulus: 3√2
  • argument: -3π/4  (or 5π/4)

Step-by-step explanation:

The modulus is the magnitude of the complex number; the argument is its angle (usually in radians).

__

<h3>rectangular form</h3>

The complex number can be cleared from the denominator by multiplying numerator and denominator by its conjugate:

  \dfrac{-9+3i}{1-2i}=\dfrac{(-9+3i)(1+2i)}{(1-2i)(1+2i)}=\dfrac{-9+3i-18i-6}{1+4}=-3-3i

<h3>polar form</h3>

The magnitude of this number is the root of the sum of the squares of the real and imaginary parts:

  modulus = √((-3)² +(-3)²) = 3√2

The argument is the arctangent of the ratio of the imaginary part to the real part, taking quadrant into consideration.

  arg = arctan(-3/-3) = -3π/4  or  5π/4 . . . . radians

__

  modulus∠argument = (3√2)∠(-3π/4)

3 0
2 years ago
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