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Citrus2011 [14]
3 years ago
11

In right triangle HJK, angle J is a right angle and tan angle H = 1. Which Statement about triangle HJK MUST BE TRUE? A. Sin ang

le H =1/2. B. Sin angle H = 1. C. Sin angle H = Cos angle H. D. Sin angle H = 1 divided by Cos angle h
Mathematics
2 answers:
NNADVOKAT [17]3 years ago
6 0

Answer:

<h2>C. sin(\angle H)=cos(\angle H)</h2>

Step-by-step explanation:

Givens

\triangle HJK is a right triangle.

m \angle J = 90\°

tan (\angle H) = 1

We this information, we can deduct that the opposite leg to \angle H is JK, and the adjacent leg to \angle H is JH. So, if  tan (\angle H) = 1, this means that both legs are equal, because to result the tangent in 1, both legs have to be equal

tan(\angle H) = \frac{JK}{JH}=1

Also, we can deduct that the angle \angle H is equal to 45°, because when the tangent is equal to one unit, that means the triangle is symmetric, which means that its angles are 45°.

So, with knowing the measure of \angle H, we can find the rest of trigonometric reasons

sin(\angle H)=sin (45\°)=\frac{\sqrt{2} }{2} \\\\cos(\angle H)=cos(45\°)=\frac{\sqrt{2} }{2}

Basically, this means that both reasons are equal

sin(\angle H)=cos(\angle H)

Therefore, the right answer is C.

Lubov Fominskaja [6]3 years ago
5 0
I hope this helps you


tgH=HJ/HK


angle H+angle K=90


H=K=45 degree


sin45=cos46 = square root of 2/2
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(a) See below

(b) r = 0.9879  

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is  

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

<u>    x   </u>    <u>      y     </u>   <u>       xy     </u>    <u>              x²    </u>   <u>       y²    </u>

   36       0.22              7.92               1296           0.05

   67        0.62            42.21              4489           0.40

   93         1.00            93.00           20164           3.46

 433        11.8          5699.4          233289        139.24

 887      29.3         25989.1          786769       858.49

1785      82.0        146370          3186225      6724

2797     163.0         455911         7823209    26569

<u>3675 </u>  <u> 248.0  </u>    <u>   911400      </u>  <u>13505625</u>   <u> 61504        </u>

9965   537.81     1545776.75  25569715   95799.63

(ii) Calculate the correlation coefficient

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}

(c) Regression line

The equation for the regression line is

y = a + bx where

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy  - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} -  \dfrac{9\times 1545776.75  - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

<u>    x    </u>   <u>      y     </u>   <u>Estimated</u>   <u>Residual </u>

    36        0.22        -10                 10

    67        0.62          -8                  9

    93        1.00           -7                  8

   142        1.86           -3                  5

  433       11.8             19               -  7

  887     29.3             45               -16  

 1785     82.0            104              -22

2797    163.0            170               -  7

3675   248.0            228               20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

4 0
3 years ago
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