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3 years ago

Describe the difference between Complementary angles and Supplementary Angles.

Mathematics

2 answers:

Archy [21]3 years ago

5 0

If two angles are complementary, then the

sum of their measures is 90 degrees.

For example, if we have a 30° angle and a 60° angle, these two angles

are complementary because the sum of their measures is 90 degrees.

If two angles are supplementary, then the

sum of their measures is 180 degrees.

For example, if we have a 100° angle and an 80° angle,

then these two angles are supplementary because

the sum of their measures is 180 degrees.

masya89 [10]3 years ago

3 0

Complementary angles equal 90 degrees and supplementary angles equal 180 degrees. It is easier to remember the complementary as 90 and to double that to get the supplementary or equal to 190 degrees. Or you could think of it in alphabetical order. C is before s and complementary angles are less that supplementary angles.

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In the function y = 5x, what is the value of x?

Sloan [31]

Answer:y=5x --> 5x=y. divide both sides with 5

Step-by-step explanation:

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3 years ago

Wich is greater, 0.04 or 0.4

neonofarm [45]

0.4 is greater because 0.04 is has a zero in front of it

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4 years ago

Read 2 more answers

If DE bisects <ABC.find the value of x if m<ABC=7x+29 and m<DBC=99°

SashulF [63]

Answer:

Step-by-step explanation:

If DE bisects <ABC, this means that <ABE = <EBC

Also <DBC+<EBC = 180

99+<EBC = 180

<EBC = 180-99

<EBC = 81

Also <ABC =<ABE+<EBC

<ABC =<EBC+<EBC

7x+29 = 81+81

7x = 162-29

7x = 133

x = 133/7

x = 19°

Hence the value of x is 19°

8 0

3 years ago

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$

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2 years ago

Distributive property of-(16-5×)

Anarel [89]

Distribute the negative to the 16 and into the -5
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3 years ago