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posledela
3 years ago
8

Why is there no tutor​

Mathematics
2 answers:
kupik [55]3 years ago
7 0
I am not sure why there isn’t any tutors
pickupchik [31]3 years ago
4 0
I think you have to follow someone and ask them to be your tutor.
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Decimal greater than 1 write the word form and expanded form for each 3.4
True [87]
3.4, 3 + 0.4, three point four
5.8, 5 + 0.8, five point eight
6.2, 6 + 0.2, six point two
5 0
3 years ago
Find the slope of the line that passes through \left(10,8\right)(10,8) and \left(1,9\right)(1,9).
sdas [7]

Answer:

m = -1/9

Step-by-step explanation:

7 0
2 years ago
there are 3 liters of apple juice at a school party. 10 students want to drink all the apple juice, Andi all want to get exactly
pochemuha

Answer:

0.3 L

Step-by-step explanation:

3 L / 10 students = 3/10 = 0.3 Liters per student

4 0
3 years ago
Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

4 0
3 years ago
. In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one stu
Sergeeva-Olga [200]

Answer: B. H_0:p_1=p_2

H_a:p_1>p_2

Step-by-step explanation:

Let p_1 be the proportion of all U.S. adults that smoked in 1995. Let p_2 denote the proportion of all U.S. adults that smoked in 2000.

Objective for the study : Whether the proportion of U.S. adults that smoke declined during the 5-year period between the samples.

i.e. p_1>p_2

Alternative hypothesis(H_a) is the statement containing population  parameter that shows there is significant difference between the groups being tested.

⇒ H_a:p_1>p_2

Null hypothesis(H_0) is the statement containing population parameter that shows there is no significant difference between the groups being tested.

⇒ H_0:p_1=p_2

Hence, The hypotheses to test in this problem are

H_0:p_1=p_2

H_a:p_1>p_2

Hence, the correct answer is B. H_0:p_1=p_2

H_a:p_1>p_2

8 0
3 years ago
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