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Gennadij [26K]
3 years ago
9

The point (1,-1) is on the terminal side of angle theta in standard position what are the values of sine cuisine and tangent of

theta
Mathematics
1 answer:
pickupchik [31]3 years ago
3 0
\bf \textit{terminal point }\implies 
\begin{array}{lclcll}
(&1&,&-1)\\
&\uparrow &&\uparrow \\
&x&&y
\end{array}
\\\\\\
sin(\theta)=\cfrac{y}{r}
\qquad 
% cosine
cos(\theta)=\cfrac{x}{r}
\qquad 
% tangent
tan(\theta)=\cfrac{y}{x}\\\\
-----------------------------\\\\
\textit{so, what's the radius "r"?}\implies r^2=x^2+y^2\implies r=\sqrt{x^2+y^2}
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Answer:

g(5)=-2(5)+1

g(5)=-10+1

g(5)=-9

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Which of the following graph represents the solution set for 5x+38 ≤ 4(2-5x)
Dvinal [7]

Answer:

Option C

Step-by-step explanation:

Given inequality is,

5x + 38 ≤ 4(2 - 5x)

5x + 38 ≤ 8 - 20x [Distributive property]

5x + 20x + 38 ≤ 8 - 20x + 20x [By adding 20x on both the sides of the inequality]

25x + 38 ≤ 8

(25x + 38) - 38 ≤ 8 - 38 [By subtracting 38 on both the sides of the inequality]

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Suppose that scores on an aptitude test are normally distributed with a mean of 100 and a standard deviation of 4.6. Scores on a
DerKrebs [107]

Answer:

<u>The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

Step-by-step explanation:

1. Let's check all the information given to us to answer the question correctly:

Mean of the scores on the aptitude test = 100

Standard deviation of the aptitude test = 4.6

Felix's score on the aptitude test = 106

Mean of the scores on the knowledge test = 70

Standard deviation of the knowledge test = 2.4

Felix's score on the knowledge test = 75

2. Which statement best describes Felix's scores on the two tests comparatively?

Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.

Therefore, a score of 106 on the aptitude test, will have the following z-score:

106 - 100 = 6 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

6/4.6 = 1.30

A score of 75 on the knowledge test, will have the following z-score:

75 - 70 = 5 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

5/2.4 = 2.08

The z-scores of Felix were 1.30 on the aptitude test and 2.08 on the knowledge test. He performed better on the aptitude test because 2.08 > 1.30, so the correct statement that best describes Felix's scores on the two tests comparatively is<u> C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

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2a(7b+5)

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