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Goryan [66]
3 years ago
11

Find the exact value of tan θ.

Mathematics
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

The answer is C.

Step-by-step explanation:

Recall SohCahToa, where tan(\theta)=opp/adj.

In the triangle, the opposite (to angle theta) is 6, while the adjacent is 2\sqrt{5}.

Put this into the equation:

tan(\theta)=\frac{6}{2\sqrt5}

Now, simplify: (multiply both top and bottom by \sqrt5

\frac{6}{2\sqrt{5} } \cdot\frac{\sqrt{5} }{\sqrt{5}} =\frac{6\sqrt{5}}{2(5)} =\frac{6\sqrt{5}}{10}=\frac{3\sqrt{5} }{5}

The answer is C.

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Find the next term is in the explicit and recursive rule for the in term of the sequence. NO LINKS!!!
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Answer:

  5) 729, an=3^n, a[1]=3; a[n]=3·a[n-1]

  6) 1792, an=7(4^(n-1)), a[1]=7; a[n]=4·a[n-1]

Step-by-step explanation:

The next term of a geometric sequence is the last term multiplied by the common ratio. (This is the basis of the recursive formula.)

The Explicit Rule is ...

  a_n=a_1\cdot r^{n-1}

for first term a₁ and common ratio r.

The Recursive Rule is ...

  a[1] = a₁

  a[n] = r·a[n-1]

__

5. First term is a₁ = 3; common ratio is r = 9/3 = 3.

Next term: 243×3 = 729

Explicit rule: an = 3·3^(n-1) = 3^n

Recursive rule: a[1] = 3; a[n] = 3·a[n-1]

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6. First term is a₁ = 7; common ratio is r = 28/7 = 4.

Next term: 448×4 = 1792

Explicit rule: an = 7·4^(n-1)

Recursive rule: a[1] = 7; a[n] = 4·a[n-1]

7 0
3 years ago
Factories the following<br> 5a + 5b
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Answer:

5(a+b)

Step-by-step explanation:

5(a)+5(b)

5(a+b)

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2 years ago
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Help solving this Math problem
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1. Formula for volume of a cone is: V = PI * r * h/3

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2.  Formula for volume of a sphere is: V = 4/3 * PI * r^3

Volume = 4/3 * PI * 1^3 = 4.19 cubic inches.

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A(1, 3), B(5, 3), and D(1, -2) are three vertices of rectangle ABCD. The coordinates of vertex C are
harkovskaia [24]
The answer is C, (5,-2) if you graph all the other points you can easily see where the missing point should be considering it's a rectangle
6 0
3 years ago
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
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