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Harlamova29_29 [7]
2 years ago
12

Write the expression as one involving only sin x and cos x. sin 2x + cos x

Mathematics
1 answer:
AfilCa [17]2 years ago
6 0

Answer:

sin 2x + cos x =  2 sin x cos x + cos x = (2 sin x + 1)cos x

Step-by-step explanation:

Given the expression: sin 2x + cos x,

then we can use the formula: sin 2x = 2 sin x cos x, which gives:

sin 2x + cos x =  2 sin x cos x + cos x = (2 sin x + 1)cos x

So there you have two expressions in terms of sin x and cos x, as requested. :D

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The image shows a geometric representation of the function f(x) = x2 – 2x – 6 written in standard form. An algebra tile configur
Over [174]

Answer:

f(x)=(x-1)^2-7

Step-by-step explanation:

Standard form:

f(x)=x^2-2x-6

Vertex form:

y=(x-h)^2+k

(h, k) is the vertex

x is the coordinate of the vertex

To find x use the formula:

x=\frac{-h}{2a}

f(x)=x^2-2x-6

a=1

b=-2

c=-6

Substitute:

x=\frac{-(-2)}{2(1)}=1

Substitute x in equation:

f(x)=1^2-2(1)-6\\f(x)=1-2-6\\f(x)=-7

h=1

k=-7

Substitute h, and k into the vertex formula:

f(x)=(x-1)^2-7

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3 years ago
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Label each of the triangles illustrated below
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A: Equilateral
B: Obtuse
C: Obtuse
D: Right
E: Obtuse
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Mary had swimming lessons every 3rd day and diving lessons every fifth day. If she had a swimming lesson and a diving lesson on
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The ratio of Monarch butterflies to Queen butterflies at a butterfly farm was 5:2. If there are no additional butterflies at the
andrew-mc [135]
So we know if there are 5 groups/dozens of Monarch butterflies, there are 2 groups/dozens of Queen butterflies. In other words, there are 5 Monarch butterflies for every 2 Queen butterflies.

Then we can turn that into an equation.

5x = m \\ \\ 2x =q \\ \\ 5x + 2x = m + q
From the last equation we wrote we can see that the total number of butterflies in the farm is 7x.

When we compare the Queen butterflies to total butterflies, we get \frac{q}{m + q} = \frac{2x}{5x + 2x} = \frac{2x}{7x} = \frac{2}{7}

The ratio of Queen butterflies to total butterflies is 2:7.
7 0
3 years ago
Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr
Minchanka [31]

Answer:

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  The margin of error is the range of values below and above the sample statistic in a confidence interval.  Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  [tex]p_1 represent the real population proportion for 1

\hat p_1 =0.768 represent the estimated proportion for 1

n_1=92 is the sample size required for 1

p_2 represent the real population proportion for 2

\hat p_2 =0.646 represent the estimated proportion for 2

n_2=95 is the sample size required for 2

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}  

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

ME= \frac{0.2753-(-0.0313)}{2}=0.1533

And we know that the margin of erro is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}

We have all the values except the value for z_{\alpha/2}

So we can find it like this:

0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}

And solving for z_{\alpha/2} we got:

z_{\alpha/2}=2.326

And we can find the value for \alpha/2 with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98 or 98%

7 0
3 years ago
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