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sineoko [7]
3 years ago
10

#1 Using the right triangle below, find the cosine of angle A.

Mathematics
1 answer:
motikmotik3 years ago
7 0

Answer: 0.8

Step-by-step explanation:

Using the Cosine formula :

Cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}

a = 6

b = 8

c = 10

substituting into the formula , we have

Cos A = \frac{8^{2}+10^{2}-6^{2}}{2(8)(10)}

Cos A = \frac{64+100-36}{160}

Cos A = \frac{164-36}{160}

Cos A = \frac{128}{160}

Therefore :

Cosine of angle A = 0.8

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Y=x-4<br> 2x+y=5<br> What’s the answer
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4 years ago
Which of the following numbers can be expressed as repeating decimals?
GalinKa [24]

Answer:

It's \frac{2}{9} and \frac{5}{6}

3 0
2 years ago
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Find the coordinates of the centroid of the triangle with the given vertices. (Lesson 5-2)
marysya [2.9K]

The triangle with vertices located at A(1 , 7), B(4 , 2), and C(7 , 7) has its centroid located at (4 , 16/3).

The centroid of a triangle is the point inside the triangle where the three medians intersect. The median of a triangle is the line the connects the midpoint of a side and the opposite vertex.

To get the centroid of a triangle, use the formula given by:

C(x , y) = [(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3]

where C is the centroid of the triangle located at (x , y)

x1, x2, and x3 are the x-coordinate of the vertices of the triangle

y1, y2, and y3 are the y-coordinate of the vertices of the triangle

Let Point 1(x1 , y1) =  A(1 , 7)

Point 2(x2 , y2) = B(4 , 2)

Point 3(x3 , y3) = C(7 , 7)

Plug in the values in the formula.

C(x , y) = [(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3]

C(x ,y) = [(1 +4 + 7)/3 , (7 + 2 + 7)/3]

C(x ,y) = (12/3 , 16/3)

C(x ,y) = (4 , 16/3)

Learn more about centroid of triangle here: brainly.com/question/7644338

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5 0
1 year ago
An indoor track is made up of a rectangular region with two semi-circles at the ends. The distance around the track is 400 meter
dybincka [34]

Answer:

width of rectangle = 2R = (200/π) = 400/π meters

length of rectangle = 400 - π(200/π) = 400 - 200 = 200 meters

Step-by-step explanation:

The distance around the track (400 m) has two parts:  one is the circumference of the circle and the other is twice the length of the rectangle.

Let L represent the length of the rectangle, and R the radius of one of the circular ends.  Then the length of the track (the distance around it) is:

Total = circumference of the circle + twice the length of the rectangle, or

         =                    2πR                    + 2L    = 400 (meters)  

This equation is a 'constraint.'  It simplifies to πR + L = 400.  This equation can be solved for R if we wish to find L first, or for L if we wish to find R first.  Solving for L, we get L = 400 - πR.

We wish to maximize the area of the rectangular region.  That area is represented by A = L·W, which is equivalent here to A = L·2R = 2RL.  We are to maximize this area by finding the correct R and L values.

We have already solved the constraint equation for L:  L = 400 - πR.  We can substitute this 400 - πR for L in

the area formula given above:    A = L·2R = 2RL = 2R)(400 - πR).  This product has the form of a quadratic:  A = 800R - 2πR².  Because the coefficient of R² is negative, the graph of this parabola opens down.  We need to find the vertex of this parabola to obtain the value of R that maximizes the area of the rectangle:        

                                                                   -b ± √(b² - 4ac)

Using the quadratic formula, we get R = ------------------------

                                                                            2a

                                                   -800 ± √(6400 - 4(0))           -1600

or, in this particular case, R = ------------------------------------- = ---------------

                                                        2(-2π)

            -800

or R = ----------- = 200/π

            -4π

and so L = 400 - πR (see work done above)

These are the dimensions that result in max area of the rectangle:

width of rectangle = 2R = (200/π) = 400/π meters

length of rectangle = 400 - π(200/π) = 400 - 200 = 200 meters

5 0
3 years ago
WILL GIVE BRAINLY PLEASE HELP!!!!!!!
marta [7]

Answer:

r² = 0.5652  < 0.7 therefore, the correlation between the variables does not imply causation

Step-by-step explanation:

The data points are;

X,          Y

0.7,       1.11

21.9,     3.69

18,         4

16.7,      3.21

18,         3.7

13.8,      1.42

18,          4

13.8,      1.42

15.5,      3.92

16.7,       3.21

The correlation between the values is given by the relation

Y =   b·X + a

b = \dfrac{N\sum XY - \left (\sum X  \right )\left (\sum Y  \right )}{N\sum X^{2} - \left (\sum X  \right )^{2}}

a = \dfrac{\sum Y - b\sum X}{N}

Where;

N = 10

∑XY = 499.354

∑X = 153.1

∑Y = 29.68

∑Y² = 100.546

∑X² = 2631.01

(∑ X)² = 23439.6

(∑ Y)² = 880.902

From which we have;

b = \dfrac{10 \times 499.354 -153.1 \times 29.68}{10 \times 2631.01 - 23439.6} = 0.1566

a = \dfrac{29.68 - 0.1566 \times 153.1}{10} = 0.5704

r = \dfrac{N\sum XY - \left (\sum X  \right )\left (\sum Y  \right )}{\sqrt{\left [N\sum X^{2} - \left (\sum X  \right )^{2} \right ]\times \left [N\sum Y^{2} - \left (\sum Y  \right )^{2} \right ]}}

r = \dfrac{10 \times 499.354 -153.1 \times 29.68}{\sqrt{\left (10 \times 2631.01 - 23439.6  \right )\times \left (10 \times 100.546- 880.902\right )}  } = 0.7518

r² = 0.5652  which is less than 0.7 therefore, there is a weak relationship between the variables, and it does not imply causation.

7 0
3 years ago
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