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Vanyuwa [196]
3 years ago
9

Prove that sin(3x)/sin(x)·cos(x) = 4cos(x)-sec(x)

Mathematics
1 answer:
MatroZZZ [7]3 years ago
8 0

Here is a huge hint:

sin(3x) = 3 sin (x) - 4(sin x)^3

sec (x) = 1/sin(x)

Take it from here.

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Which point is an x-intercept of the quadratic function<br> f(x) = (x + 6)(x – 3)?
Scilla [17]
Set each factor equal to zero and solve.
x + 6 = 0 x - 3 = 0
x = - 6 x = 3
7 0
3 years ago
Read 2 more answers
A sample of an unknown liquid has a volume of 12.0 mL and a mass of 6 g. What is its density?
NemiM [27]
Density = Mass/Volume
Density = 6/12 = 0.5 g/mL or 0.5 kg/L

6 0
3 years ago
Plz answer DDDDDDDDDDDDDDDDDD! using the line plot
masya89 [10]

Answer:

a. 6

b. 9

c. 4

Step-by-step explanation:

a.

Range = greatest - smallest

Range = 85 - 79

Range = 6

b.

b > 81

82, 83, 84...

3 82s

4 83s

2 85s

3 + 4 + 2 = 9

c.

b = 83

4 83s

4

6 0
3 years ago
Assuming the amount of money college students spend on text books each semester is symmetrical with a mean of 500 and a standard
TiliK225 [7]

16% percent of the students paid MORE than Jane.

The mean(μ) of money spent on textbooks is 500

The standard deviation(σ) of money spent on textbooks is 50

Money paid by Jane for her books is $550

We will use this formula,

Ζ=x-μ/σ

To find: the percentage of students paid MORE than Jane for the textbooks

P(X > 550)=?

Solution:

P(X > 550)=1-P(X≤550)

=1-P(Ζ≤\frac{550-500}{50} )

=1-P(Ζ≤ 1)

=1-0.8413

=0.1587

≈16%

Therefore, 16% percent (approx) of the students paid MORE than Jane.

Learn more about mean and standard deviation here brainly.com/question/4388715

#SPJ4

3 0
2 years ago
I need help!!<br> I don’t get it!!
Len [333]

Answer:

170

Step-by-step explanation:

3 0
3 years ago
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