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3 years ago

Find the real zeros of the trigonometric function on the interval 0 ≤ x < 2π

Mathematics

2 answers:

Luda [366]3 years ago

6 0

4cos2(x)−3=0

or
4cos2(x)=3

cos(x)= √3/2

π/6, 5π/6, 7π/6, 11π/6

Romashka-Z-Leto [24]3 years ago

5 0

F(x) = 4 [cos (x)]^2 - 3 = 0

4[cos(x)]^2 = 3

cos(x) = √3 / 2

That happens in the first and fourth quadrants, for the angles 30 degrees and 330 degrees.

Answer: x = 30 degrees and x = 330 degrees

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Which ordered pairs are solutions to the inequality 2y-x≤-6 ?

Ray Of Light [21]

Answer:

Step-by-step explanation:

3 0

2 years ago

Complete the proofs using the most appropriate method. may require CPCTC.

n200080 [17]

For the first one, you did good. I will just suggest a couple things.

Statement Reason

JK ≅ LM Given

<JKM ≅ < LMK Given (You did both of these steps so well done.)

MK ≅ MK Reflexive Property (Your angle pair is congruent but isn't one of the interior angle of the triangles you are trying to prove.)

ΔJMK ≅ ΔLKM SAS

Problem 2: (You also have a lot of great stuff here.)

Statement Reason

DE ║ FG Given

DE ≅ FG Given

<DEF≅<FGH Given

<EDF≅<GFH Corresponding Angles (You don't need to know that F is the midpoint but you got corresponding angle pair which is correct.)

ΔEDF≅ΔGFH ASA

<DFE≅<FHG CPCTC

3 0

3 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

2 years ago

Can someone help plz

Ivenika [448]

Answer:

X=27

Step-by-step explanation:

$\frac{24}{32} = \frac{30}{40} = \frac{x}{36}$

so x=27

4 0

2 years ago

Which of the following is the probabilty against drawing a heart from a standard deck of 52 cards?

nevsk [136]

<span>p = 39/52 = 3/4 = 0.75
</span>
13 cards each of 4 different suits
if you draw a card, there are 39 cards that
are not hearts out of 52 all together.

3 0

3 years ago