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3 years ago

Find all angles in [0,2pi) that satisfy the equation: 3csc^2()cot(x)=−4cot(x)

Mathematics

1 answer:

MrRissso [65]3 years ago

5 0

The answer is 20 because w X 4 is 20

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Anyoneee help im stuck !! 15 points!!

Gnom [1K]

Answer:

Given: $\angle 1$ , $\angle 2$ , $\angle 3$ , $\angle 4$ formed by two intersecting segments.

In the given figure;

Linear pair states that a pair adjacent angle formed when two lines intersect.

Then by definition of linear pairs,

$\angle 1$ and $\angle 2$ forms a linear pair

Also, $\angle 2$ and $\angle 3$ forms a linear pair.

Linear pair postulates states that the two angle that forms a linear pair are supplementary(i,e add up to 180 degree).

Then by linear pair postulates;

$m\angle 1+ m\angle 2 =180^{\circ}$

and

$m\angle 2+m \angle 3 =180^{\circ}$

Substitution property of equality states that if x =y then, x can be substituted in for y or vice -versa.

then by substitution property of equality:

$m\angle 1+m\angle 2=m\angle 2+ m\angle 3$

Addition property of equality states that:

if x =y, then x + z = y+ z

By addition property of equality:

$m\angle 1 =m\angle 3$ hence proved!

6 0

3 years ago

Brad had a small gathering at a local steakhouse. The steakhouse offers three dinner platters which vary by size and price. They

Naya [18.7K]

9.95×4=39.80
12.95×2=25.90
15.95×2=31.90
39.80+ 25.90+31.90=97.60
97.60×.18=17.56
97.60+17.56=$115.16 without tax and 18% gratutity

4 0

2 years ago

What is the center and radius of the circle?

riadik2000 [5.3K]

Answer:

Center = (5,-3) radius = sqrt (36) = 6

Step-by-step explanation:

4 0

2 years ago

Find the x- and y-intercepts of the graph of -x + 3y = 37. State your answers as

kifflom [539]

Answer:

Hello!

The answer to this question is:

X = -37

y = -37/3

Step-by-step explanation:

6 0

2 years ago

Yuri thinks that 3/4 is a root of the following function. q(x) = 6x3 + 19x2 – 15x – 28 Explain to Yuri why 3/4 cannot be a root.

k0ka [10]

Answer:

Yuri is not correct.

Step-by-step explanation:

Given expression is q(x) = 6x³ + 19x² - 15x - 28

If 'a' is a root of the given function, then by substituting x = a in the expression, q(a) = 0

Similarly, for x = $\frac{3}{4}$ ,

$q(\frac{3}{4})=6(\frac{3}{4})^3+19(\frac{3}{4})^2-15(\frac{3}{4})-28$

= $6(\frac{27}{64})+19(\frac{9}{16})-15(\frac{3}{4})-28$

= $(\frac{162}{64})+(\frac{171}{16})-(\frac{45}{4})-28$

= $(\frac{162}{64})+(\frac{684}{64})-(\frac{720}{64})-\frac{1792}{64}$

= $-\frac{1666}{64}$

= $-\frac{833}{32}$ ≠ 0

Therefore, Yuri is not correct. x = $\frac{3}{4}$ can not be a root of the given expression.

6 0

3 years ago

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