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netineya [11]
3 years ago
15

Adjacent, right angles are complementary. always, sometimes, or never

Mathematics
2 answers:
IgorLugansk [536]3 years ago
7 0
Adjacent right angles are Never complementary
DerKrebs [107]3 years ago
4 0
What you'll need to answer this question are the definitions of adjacent, right angle, and complementary. For the sake of this question, I'll give you those definitions:

- Two angles are <em>adjacent</em> when they lie next to each other across a line. Imagine you have two line segments forming an angle, and that that angle is cut in half by another line segment, forming two smaller angles. Each of those smaller angles is <em>adjacent</em> to the other.
- A <em>right </em>angle is an angle formed at the meeting point of two <em>perpendicular </em>lines. For an example of what it means to be perpendicular, think of the angles a plus sign forms at each of its corners - those lines are what are called perpendicular.
- <em>Complementary</em> angles are two angles which add up to a right angle.

With this knowledge, take a look at the question again. Is it possible for two <em>right angles</em> that are <em>next to each other</em> to add up to <em>another right angle?</em> Can <em>any </em>number greater than zero, when added to itself, ever <em>equal itself</em>?
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Please help me find the area of shaded region and step by step​
Bumek [7]

Answer:

Part 1) A=60\ ft^2

Part 2) A=80\ cm^2

Part 3) A=96\ m^2

Part 4) A=144\ cm^2

Part 5) A=9\ m^2

Part 6) A=(49\pi -33)\ in^2

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

A=bh

where

b is the base of rectangle

h is the height of rectangle

so

A=(12)(7)-(8)(3)

A=84-24

A=60\ ft^2

Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

A=b^2

where

b is the length side of the square

so

A=(12)(8)-(4^2)

A=96-16

A=80\ cm^2

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

so

A=4(4)(5)+(4^2)

A=80+16

A=96\ m^2

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

so

A=(15^2)-(9^2)

A=225-81

A=144\ cm^2

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

A=\frac{1}{2}(b)(h)

where

b is the base of triangle

h is the height of triangle

so

A=\frac{1}{2}(6)(7)-(6)(2)

A=21-12

A=9\ m^2

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

A=\pi r^{2}

where

r is the radius of the circle

so

A=\pi (7^2)-(3)(11)

A=(49\pi -33)\ in^2

7 0
3 years ago
What's the answer for <br> x/2-5=3
kifflom [539]
X=16


I use Tiger Algebra for help and it gives me all the correct work and answers as i need them!! TRY USING IT TODAY!!!!! 
4 0
3 years ago
Which of the following quartic functions has x=-1 and x=-2 as its only two real zeros
Brums [2.3K]
Quartic is 4th degree
the factors of an equation with roots r1,r2 is
(x-r1)(x-r2)
4th degree
it could be
(x-r1)¹(x-r2)³ or
(x-r1)²(x-r2)² or
(x-r1)³(x-r2)¹


roots or zeroes at x=-1 and x=-2
(x-(-1)) and (x-(-2))
(x+1) and (x+2)

the function could be factored into
(x+1)¹(x+2)³ or
(x+1)²(x+2)² or
(x+1)³(x+2)¹

expanded would be
x⁴+7x³+18x²+20x+9 or
x⁴+6x³+13x²+12x+4 or
x⁴+5x³+9x²+7x+2
one of those is the answer

4 0
3 years ago
ILL BRAINLIEST YOU IF YOU GET IT RIGHT
IRINA_888 [86]

Answer:

b hope it helps make brainlliest ty

5 0
3 years ago
Read 2 more answers
How many solutions does the equation 3x − 7 = 4 + 6 + 4x have? Two Zero Infinitely many One
sleet_krkn [62]

Answer:

one solution

Step-by-step explanation:

3x − 7 = 4 + 6 + 4x

Combine like terms

3x-7=4x+10

Subtract 3x from each side

-7=x+10

Subtract 10 from each side

-17 =x

There is one solution

4 0
3 years ago
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