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Veronika [31]
3 years ago
10

Find all solutions in the interval [0,2pi). cos 2x + sqrt(2) sinx=1

Mathematics
1 answer:
kykrilka [37]3 years ago
6 0
<span>cos 2x + sqrt(2) sinx=1
</span><span>
Note that: cos 2x = cos^2x - sin^2x = (1-sin^2x) - sin^2x = 1 - 2sin^2x.
So, when alternatively written, you have the following equation:

</span>- 2sin^2x + sqrt(2)sinx + 1 = 1
- 2sin^2x + sqrt(2)sinx = 0

Then, let z=sin(x). So you get,

- 2z^2 + sqrt(2)z = 0
z(- 2z + sqrt(2)) = 0

Either z=0, or - 2z + sqrt(2) = 0 --->  z=sqrt(2)/2.
Then, since z=0 or z=sqrt(2)/2, therefore sin(x)=0, or sin(x)=sqrt(2)/2.

Then, for you remains just to list the angles. (Let me know if this is not fair or if you got questions.)
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Is -3.45 closer to -3 or -4 on a number line please i really need help
oksano4ka [1.4K]

Answer:

<u>-3.45</u> is closer to -3

Step-by-step explanation:

because, since it's negative the closer the negative number is closer to 0 its number is lower (-) like -4 is farther away from the 0 than -3.

5 0
2 years ago
Geometric sequences HELP ASAP!
Pani-rosa [81]

Given:

The table for a geometric sequence.

To find:

The formula for the given sequence and the 10th term of the sequence.

Solution:

In the given geometric sequence, the first term is 1120 and the common ratio is:

r=\dfrac{a_2}{a_1}

r=\dfrac{560}{1120}

r=0.5

The nth term of a geometric sequence is:

a_n=ar^{n-1}

Where a is the first term and r is the common ratio.

Putting a=1120, r=0.5, we get

a_n=1120(0.5)^{n-1}

Therefore, the required formula for the given sequence is a_n=1120(0.5)^{n-1}.

We need to find the 10th term of the given sequence. So, substituting n=10 in the above formula.

a_{10}=1120(0.5)^{10-1}

a_{10}=1120(0.5)^{9}

a_{10}=1120(0.001953125)

a_{10}=2.1875

Therefore, the 10th term of the given sequence is 2.1875.

6 0
3 years ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)

Applying the following property of exponentials numbers in (II):

e^{a}.e^{b}=e^{a+b}

Therefore e^{-28k} can be written as e^{-14k}.e^{-14k}

152=190-Ae^{-14k}.e^{14k}

Replacing (I) in the previous equation:

152=190-76e^{-14k}

Solving for k:

Subtracting 190 both sides, dividing by -76:

0.5=e^{-14k}

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76

Dividing by 0.5

A=152

7 0
3 years ago
Find the x- and y-intercepts of the equation.<br> 3x − 3y = 12
Romashka [77]
X=-4-y there you go enjoy

6 0
3 years ago
Enter a number into the blank to make the equation true.
yKpoI14uk [10]

Answer:

779

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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