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levacccp [35]
1 year ago
10

Find the length of side AB Give answer to 3 significant figures

Mathematics
1 answer:
Rina8888 [55]1 year ago
6 0

Answer:

AB = 8.857 cm

Step-by-step explanation:

Here, we are given a <em>right angle</em> \triangle ABC in which we have the following things:

\angle A = 90 ^\circ\\\angle C = 41 ^\circ\\\text{Side }BC = 13.5 cm

Side <em>BC </em>is the hypotenuse here.

We have to find the side <em>AB</em>.

Trigonometric functions can be helpful to find the value of Side AB here.

Calculating \angle B:

Sum of all the angles in \triangle ABC is 180^\circ.

\Rightarrow \angle A + \angle B + \angle C = 180^\circ\\\Rightarrow 90^\circ + \angle B + 41^\circ = 180^\circ\\\Rightarrow \angle B = 49^\circ

We know that <em>cosine </em>of an angle is:

cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}\\\Rightarrow cos B = \dfrac{AB}{BC}\\\Rightarrow cos 49^\circ = \dfrac{AB}{13.5}\\\Rightarrow AB = 13.5 \times 0.656\\\Rightarrow AB = 8.857 cm

So, side AB = 8.857 cm .

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Todd had 6 gallons of gasoline in his motorbike. After driving 150​ miles, he had 3 gallons left. Compute the​ slope, or rate of
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What is the range of the exponential function shown below
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f(x) is an exponential function

f(x) = 9*2^x

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Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of
Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2  are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j

The unit vector in the direction of force will be

\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}

The magnitude of the force is 40lb, so the force will be

F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k

\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}

The magnitude of the force is 60lb, so the force will be

F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)

The resultant of the two forces is

F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k

The magnitude force will be

|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469

The angle formed by F and y axis

\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=

The angle formed by F and z axis

\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359

8 0
1 year ago
Find the correlation coefficient of the line of best fit for the points (-3,-40), (1,12), (5,72), (7,137). Explain how you guy y
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Answer:

r = 0.9825; good correlation.

Step-by-step explanation:

One formula for the correlation coefficient is  

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

The calculation is not difficult, but it is tedious.

1. Calculate the intermediate numbers

We can display them in a table.

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

      -3   -40    120     9     1600

       1      12      12      1        144

       5    72   360   25      5184

     <u>  7</u>   <u>137</u>  <u> 959</u>   <u>49</u>    <u>18769 </u>

Σ = 10   181  1451   84   25697

2. Calculate the correlation coefficient

r = \dfrac{n\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{4\times 1451 - 10\times 181}{\sqrt{[4\times 84 - 10^{2}][4\times25697 - 181^{2}]}}\\\\= \dfrac{5804 - 1810}{\sqrt{[336 - 100][102788 - 32761]}}\\\\= \dfrac{3994}{\sqrt{236\times70027}}\\\\= \dfrac{3994}{\sqrt{16526372}}\\\\= \dfrac{3994}{4065}\\\\= \mathbf{0.9825}

The closer the value of r is to +1 or -1, the better the correlation is. The values of x and y are highly correlated.

3 0
1 year ago
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