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Westkost [7]
4 years ago
5

two lines meet at a point that is also the endpoint of two rays. setup and solve the appropriate equations to solve for the valu

es of angles x and y. show the steps you followed to find the unknown angles.

Mathematics
2 answers:
NeTakaya4 years ago
8 0

Answer:

Step-by-step explanation:

Check attachment for labeling,

To find x,

Since angle on a straight line is 180°,

Then, AOE is a straight line

<AOE = 180°

NOTE: <AOC is a right angle,

Then, <AOC = 90°

So,

<AOE = <AOC + <COE

180 = 90 + <COE

<COE = 180-90

<COE = 90°

Then, <COE is made of two angles

<COD = 71° and <DOE = x

<COE = <COD + <DOE

90 = 71 + x

x = 90 - 71

x = 19°

Also, to find y

NOTE: BOF is a straight line and angle on a straight line is 180°

<BOF = <BOC + <COD + <DOE + <EOF

180 = 37 + 71 + x + y

180 = 108 + 19 + y

180 = 127 + y

Then, subtract 127 from both sides

y = 180 - 127

y = 53°

So, x = 19° and y = 53°

kherson [118]4 years ago
3 0

Answer:

x = 19°

y = 53°

Step-by-step explanation:

First we can find the value of x, then the value of y.

For x, we can observe from the figure that x is complementary to the angle of 71°, that is, they both summed have 90°.

So, we have that:

x + 71 = 90

x = 19°

Now we can solve for y.

From the figure we have that y is supplementary to the angle (37° + 71° + x°), that is, they summed form a angle of 180°.

So we have that:

y + 37 + 71 + 19 = 180

y = 180 - 37 - 71 - 19 = 53°

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In 2 minutes, Nolan unloads 10 boxes from the truck. Write the equation for the relationship between x and y.
Anna11 [10]

Answer:

2x = y

Step-by-step explanation:

x = minutes

y = 10 boxes

4 0
3 years ago
Find the measure of each side indicated round to the nearest 10th
earnstyle [38]

Answer:

20) 30°; 21) 21,1; 22) 6,2

Step-by-step explanation:

For the first image, you have to do <em>csc</em><em>⁻</em><em>¹</em><em> </em><em>2</em><em> </em>[OR <em>sin</em><em>⁻</em><em>¹</em><em> </em><em>½</em>] because you are solving for an angle measure. When evaluated, you get 30°.

For the second image, you have to do <em>13sec</em><em> </em><em>52</em><em>°</em><em> </em><em>=</em><em> </em><em>x</em><em> </em>[OR <em>13\</em><em>cos</em><em> </em><em>52</em><em>°</em><em> </em><em>= x</em>]. When evaluated, you get an approximate measure of 21,1.

For the third image, you have to do <em>6csc</em><em> </em><em>75</em><em>°</em><em> </em><em>=</em><em> </em><em>x</em><em> </em>[OR <em>6</em><em>\</em><em>sin</em><em> </em><em>75</em><em>°</em><em> </em><em>= x</em>]. When evaluated, you get an approximate measure of 6,2.

Extended information on Trigonometric Ratios

O\H = sin θ

A\H = cos θ

O\A = tan θ

H\A = sec θ

H\O = csc θ

A\O = cot θ

I am joyous to assist you anytime.

4 0
3 years ago
5. Find the vertex and length of the latus rectum for the parabola.
olga55 [171]

Answer: (2,5)

Step-by-step explanation:

To find the x coordinate of the vertex, you will set  x-2=0 and solve for x

x-2=0

 +2  +2

x = 2  

Now since we know the vertex has an x coordinate of 2 we have to solve for y by plotting in the x value.

y= 1/2(2-2)^2 + 5

y = 0 +5

y =5

(2,5)

5 0
3 years ago
Explain how to solve 5x^2-3x=25 by completing the square. What are the solutions?
Mila [183]

Answer:

x1 = √(509) /100 + 3/10

x2  = -√(509) /100 + 3/10

Step-by-step explanation:

We do completing the square as follows:

1. Put all terms with variable on one side and the constants on the other side.

5x^2 - 3x = 25

2. Factor out the coefficient of the x^2 term.

5(x^2 - 3x/5) = 25

3. Inside the parentheses, we add a number that would complete the square and also add this to the other side of the equation. In this case we add 9/100 and on the other  side we add 9/20.

5(x^2 - 3x/5 + 9/100) = 25 + 9/20

4. We simplify as follows:

5(x^2 - 3x/5 + 9/100) = 25 + 9/20

5(x - 3/10)^2 = 509/20

(x - 3/10)^2 = 509/100

x1 = √(509) /100 + 3/10

x2  = -√(509) /100 + 3/10

6 0
3 years ago
Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea
Gennadij [26K]

Answer:

y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ]

Step-by-step explanation:

Solution:-

- Given is the 2nd order linear ODE as follows:

                      ty'' + ( 2t - 1 )*y' - 2y = 6t^2 . e^(^-^2^t^)

- The complementary two independent solution to the homogeneous 2nd order linear ODE are given as follows:

                     y_1(t) = 2t - 1\\\\y_2 (t ) = e^-^2^t

- The particular solution ( yp ) to the non-homogeneous 2nd order linear ODE is expressed as:

                    y_p(t) = u_1(t)*y_1(t) + u_2(t)*y_2(t)

Where,

              u_1(t) , u_2(t) are linearly independent functions of parameter ( t )

- To determine [  u_1(t) , u_2(t) ], we will employ the use of wronskian ( W ).

- The functions [u_1(t) , u_2(t) ] are defined as:

                       u_1(t) = - \int {\frac{F(t). y_2(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\\\u_2(t) =  \int {\frac{F(t). y_1(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\

Where,

      F(t): Non-homogeneous part of the ODE

      W [ y1(t) , y2(t) ]: the wronskian of independent complementary solutions

- To compute the wronskian W [ y1(t) , y2(t) ] we will follow the procedure to find the determinant of the matrix below:

                      W [ y_1 ( t ) , y_2(t) ] = | \left[\begin{array}{cc}y_1(t)&y_2(t)\\y'_1(t)&y'_2(t)\end{array}\right] |

                      W [ (2t-1) , (e^-^2^t) ] = | \left[\begin{array}{cc}2t - 1&e^-^2^t\\2&-2e^-^2^t\end{array}\right] |\\\\W [ (2t-1) , (e^-^2^t) ]= [ (2t - 1 ) * (-2e^-^2^t) - ( e^-^2^t ) * (2 ) ]\\\\W [ (2t-1) , (e^-^2^t) ] = [ -4t*e^-^2^t ]\\

- Now we will evaluate function. Using the relation given for u1(t) we have:

                     u_1 (t ) = - \int {\frac{6t^2*e^(^-^2^t^) . ( e^-^2^t)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_1 (t ) =  \frac{3}{2} \int [ t*e^(^-^2^t^) ] \, dt\\\\u_1 (t ) =  \frac{3}{2}* [ ( -\frac{1}{2} t*e^(^-^2^t^) - \int {( -\frac{1}{2}*e^(^-^2^t^) )} \, dt]  \\\\u_1 (t ) =  -e^(^-^2^t^)* [ ( \frac{3}{4} t +  \frac{3}{8} )]  \\\\

- Similarly for the function u2(t):

                     u_2 (t ) =  \int {\frac{6t^2*e^(^-^2^t^) . ( 2t-1)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_2 (t ) =  -\frac{3}{2} \int [2t^2 -t ] \, dt\\\\u_2 (t ) =  -\frac{3}{2}* [\frac{2}{3}t^3 - \frac{1}{2}t^2  ]  \\\\u_2 (t ) =  t^2 [\frac{3}{4} - t ]

- We can now express the particular solution ( yp ) in the form expressed initially:

                    y_p(t) =  -e^(^-^2^t^)* [\frac{3}{2}t^2 + \frac{3}{4}t - \frac{3}{8} ]    + e^(^-^2^t^)*[\frac{3}{4}t^2 - t^3 ]\\\\y_p(t) =  -e^(^-^2^t^)* [t^3 + \frac{3}{4}t^2 + \frac{3}{4}t - \frac{3}{8} ] \\

Where the term: 3/8 e^(-2t) is common to both complementary and particular solution; hence, dependent term is excluded from general solution.

- The general solution is the superposition of complementary and particular solution as follows:

                    y_g(t) = y_c(t) + y_p(t)\\\\y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ]

                   

3 0
3 years ago
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