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valentina_108 [34]
3 years ago
7

In Which Quadrant is this true

Mathematics
1 answer:
34kurt3 years ago
5 0

Given:

\sin \theta

\tan \theta

To find:

The quadrant in which \theta lie.

Solution:

Quadrant concept:

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II, only \sin\theta and \csc\theta are positive.

In Quadrant III, only \tan\theta and \cot\theta are positive.

In Quadrant IV, only \cos\theta and \sec\theta are positive.

We have,

\sin \theta

\tan \theta

Here, \sin\theta is negative and \tan\theta is also negative. It is possible, if \theta lies in the Quadrant IV.

Therefore, the correct option is D.

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2x+4 is the length of one side
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HELP ITS AN EMERGENCY
forsale [732]

Answer:

<u>A. Vertex: (1, -8); x-intercepts: -1 and 3</u>

Step-by-step explanation:

The vertex is the point of the parabola. Basically, the part where the parabola 'scoops' up.

So the point where it scoops up is (1, -18)

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The places where it touches the x-axis is -1 and 3

7 0
3 years ago
BRAINLIEST !! Evaluate the following fractions giving your awnser in its simplest form.
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Answer:

question no c 0•08

Step-by-step explanation:

at first

1/4+1/3-1/2

0•25+0.33-0.50

0.58-0.50

0.08 answer

7 0
1 year ago
Read 2 more answers
1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
Ad libitum [116K]
Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21
5 0
3 years ago
How many whole sandwiches can be made with 2 pounds of cheese if each sandwich has 3 ounces of cheese on it?
Sidana [21]
2 pounds  =32 ounces divide that by 3 its equals about 10 sandwiches
4 0
3 years ago
Read 2 more answers
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