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anyanavicka [17]
2 years ago
11

What is the quotient of StartStartFraction 90 (cosine (StartFraction pi Over 4 EndFraction) + I sine (StartFraction pi Over 4 En

dFraction) ) OverOver 2 (cosine (StartFraction pi over 12 EndFraction) + I sine (StartFraction pi Over 12 EndFraction) ) EndEndFraction ?
45 (cosine (StartFraction pi Over 3 EndFraction) + I sine (StartFraction pi Over 3 EndFraction) )

88 (cosine (StartFraction pi Over 3 EndFraction) + I sine (StartFraction pi Over 3 EndFraction) )

45 (cosine (StartFraction pi Over 6 EndFraction) + I sine (StartFraction pi Over 6 EndFraction) )

88 (cosine (StartFraction pi Over 6 EndFraction) + I sine (StartFraction pi Over 6 EndFraction) )


I think it's c/the third option
Mathematics
2 answers:
Gelneren [198K]2 years ago
8 0

Answer:

The answer is C

Step-by-step explanation:

7nadin3 [17]2 years ago
6 0

Answer:

I can confirm, C is the correct answer.

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Help me please thank you
Lapatulllka [165]
All of them except TR and PH
3 0
2 years ago
What is the solution to 2 log Subscript 9 Baseline (x) = log Subscript 9 Baseline 8 + log Subscript 9 Baseline (x minus 2) x = n
Kaylis [27]

Answer:

\large\boxed{\large\boxed{x=4}}

Step-by-step explanation:

The equation to solve is:

            2\log_9 x=\log_9 8+\log_9 (x-2)

1. <u>On the left-hand side </u>use: "The product of a constant by a logarithm is equal to the logarithm raised to the constant"

Thus, the left-hand side is:

               2\log_9 x=\log_9 x^2

2. On the <u>right-hand side</u> use "The sum of two logarithms with the same base is the logarithm of the product":

Then, on the right-hand side:

        \log_9 8+\log_9 (x-2)=\log_9 8(x-2)

3. <u>Make them equal</u>:

      \log_9 x^2=\log_9 8(x-2)

4. Since the two functions are the same, <u>make the arguments equal</u>:

     x^2=8(x-2)

5. <u>Solve the equation</u>:

        x^2=8x-16\\\\x^2-8x+16=0\\\\(x-4)^2=0\\\\x-4=0\\\\x=4\leftarrow solution

4 0
2 years ago
Read 2 more answers
1)Given: AB = 4 AD = 6
Vaselesa [24]
<span>1)Given: AB = 4 AD = 6
What is the name of the radius of the larger circle?
the answer part 1) is 
the radius </span>of the larger circle is AD

<span>2)Given: AB = 4 AD = 6
What point is in the interior of both circles?
the answer Part 2) is 
The point A  (the center of the circles)

</span><span>3) Given: AB = 4 AD = 6
Which points are in the exterior of both circles?</span><span>
the answer Part 3) is
</span><span>E and G

</span><span>4)The circles are _____.
</span><span>the answer Part 4) is
</span><span>concentric
</span>
<span>5)If AC = 20 and BD = 8, what is the radius of the smaller circle?
</span>we know that
radius smaller circle=AB
and
AB=AC-BD--------> AC=20-12-------> AC=8 units

the answer part 5) is
the radius of the smaller circle is 12 units

<span>6)Given: AB = 4 AD= 6
What is the length of BD?</span>
we know that
AD=AB+BD
solve for BD
BD=AD-AB--------> BD=6-4-----> BD=2 units

the answer Part 6) is
the length of BD is 2

<span>7)Given: AB = 4 AD = 6
What is the name of the radius of the smaller circle?</span>

the answer Part 7) is 
the name of the radius of the smaller circle is AB
5 0
2 years ago
20% of the tickets sold at a water park were adult tickets. If the park sold 55 tickets in all,how many did it sell?
djyliett [7]

Answer:

11 adult tickets

Step-by-step explanation:

I guess you want the number of adult tickets sold.

That is 20% of 55

= 0.20*55

= 11 (answer)

8 0
2 years ago
Read 2 more answers
A large restaurant is being sued for age discrimination because 15% of newly hired candidates are between the ages of 30 years a
Fudgin [204]

Answer:

See explanation below.

Step-by-step explanation:

Let's take P as the proportion of new candidates between 30 years and 50 years

A) The null and alternative hypotheses:

H0 : p = 0.5

H1: p < 0.5

b) Type I error, is an error whereby the null hypothesis, H0 is rejected although it is true. Here, the type I error will be to conclude that there was age discrimination in the hiring process, whereas it was fair and random.

ie, H0: p = 0.5, then H0 is rejected.

7 0
2 years ago
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