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xz_007 [3.2K]
3 years ago
7

In the diagram below, BD is parallel to XY. What is the value of x?​

Mathematics
2 answers:
alexira [117]3 years ago
3 0

Answer:

A. 55

Step-by-step explanation:

dybincka [34]3 years ago
3 0

Answer: A. 55

Step-by-step explanation:

You might be interested in
Pls help me quick !! Just simple quick answers
Aleksandr-060686 [28]

Answer:

j=3 inches

Step-by-step explanation:

Since the two triangles are similar, we can set up a ratio

1in:3in

j:9in

Now we solve for j

Since the length of the side of the shorter triangle is 1/3 the length of the side of the longer triangle (as shown by the ratio 1:3 which can also be written as 1/3), we can multiply 9 by 1/3. This gives us 3.

So j=3 inches

4 0
2 years ago
Which graph represents the solution set to the system of inequalities?
vekshin1

Answer:

c

Step-by-step explanation:

y-6<0,y<6

all graphs satisfy it

y<x+4

when x=0,y<4

c satisfies it.

y+6>-3x

when x=0,y>-6

c satisfies it.

a b

c d

3 0
2 years ago
Read 2 more answers
1. Se dispara una bala de 10grbcon una velocidad de 500m/s contra un muro de 10cm de espesor. Si la resistencia del muro al avan
vodomira [7]

Answer:

1) La velocidad de la bala después de atravesar el muro es de aproximadamente 435,890 metros por segundo.

2) La potencia del automóvil es 96438,272 watts o 131,208 caballos de vapor.

3) La velocidad del objeto al llegar al suelo es aproximadamente 14,005 metros por segundo.

Step-by-step explanation:

1) La velocidad final de la bala puede determinarse mediante el Teorema del Trabajo y la Energía, a partir del cual se tiene la siguiente fórmula:

\frac{1}{2}\cdot m\cdot v_{o}^{2} -F\cdot \Delta s = \frac{1}{2}\cdot m \cdot v_{f}^{2} (1)

Where:

m - Masa de la bala, en kilogramos.

v_{o}, v_{f} - Velocidades inicial y final de la bala, en metros por segundo.

F - Resistencia del muro al avance de la bala, en newtons.

\Delta s - Espesor del muro, en metros.

Si sabemos que m = 0,01\,kg, v_{o} = 500\,\frac{m}{s}, F = 3000\,N and \Delta s = 0,1\,m, entonces la velocidad final de la bala es:

v_{f}^{2}=v_{o}^{2} -\frac{2\cdot F\cdot \Delta s}{m}

v_{f} = \sqrt{v_{o}^{2}-\frac{2\cdot F\cdot \Delta s}{m} }

v_{f} \approx 435,890\,\frac{m}{s}

La velocidad de la bala después de atravesar el muro es de aproximadamente 435,890 metros por segundo.

2) Asumamos que el automóvil acelera a tasa constante, significando que la fuerza neta será constante. Para un sistema cuya fuerza neta sea constante, la potencia experimentada queda descrita por la siguiente ecuación:

P = m\cdot a(t)\cdot v(t) (2)

a(t) = a (3)

v(t) = v_{o} + a\cdot t (4)

Donde:

P - Potencia, en watts.

m - Masa del automóvil, en kilogramos.

a(t) - Aceleración, en metros por segundo al cuadrado.

v(t) - Velocidad, en metros por segundo.

v_{o} - Velocidad inicial del automóvil, en metros por segundo.

Si sabemos que m = 1000\,kg, a = 3,472\,\frac{m}{s}, v_{o} = 0\,\frac{m}{s} y t = 8\,s entonces la potencia experimentada por el automóvil es:

P = 96438,272\,W (131,208\,C.V.)

La potencia del automóvil es 96438,272 watts o 131,208 caballos de vapor.

3) El cuerpo experimenta un Movimiento de Caída Libre, el cual es un Movimiento Uniformemente Acelerado debido a la gravedad terrestre. La velocidad del cuerpo al llegar al suelo se determina mediante la siguiente fórmula cinemática:

v_{f} = \sqrt{v_{o}^{2}+2\cdot g\cdot h} (5)

Donde:

v_{o} - Velocidad inicial del cuerpo, en metros por segundo.

v_{f} - Velocidad final del cuerpo, en metros por segundo.

g - Aceleración gravitacional, en metros por segundo al cuadrado.

h - Altura recorrida por el cuerpo, en metros.

Si sabemos que v_{o} = 0\,\frac{m}{s}, g = 9,807\,\frac{m}{s^{2}} y h = 10\,m, entonces la velocidad al llegar al suelo es:

v_{f} \approx 14,005\,\frac{m}{s}

La velocidad del objeto al llegar al suelo es aproximadamente 14,005 metros por segundo.

4 0
2 years ago
Check each equation whose graph is the line that
yanalaym [24]

Answer:

Points P ( 4 , - 7 ) and  Q ( 1 , 5 ) belong to the equations:

1  ;  4  ;  and  5

Step-by-step explanation:

Equations 1 ; 4 and 5 are the same equation

Equation 1    y  =  - 4*x + 9

Equation 4

y + 7 = - 4 * ( x - 4 )   ⇒  y + 7  = - 4*x + 16   ⇒ y = - 4*x - 7 + 16

y = -4*x + 9

Equation 5

4*x + y = 9    ⇒ y = - 4*x + 9

Now for the equation y = - 4*x + 9

P ( 4 , -7)

For x = 4    y = - 4*(4) + 9     ⇒  y = - 16 +  9    ⇒  y = - 7

Then point P is in the line  y = - 4*x + 9

Point Q (1 , 5 )

For x = 1    y = - 4 * ( 1) + 9     ⇒  y = - 4 + 9    ⇒  y = 5

Point Q is in the line y = - 4*x + 9

Equation 2

y = - 4*x - 23

Point P ( 4 , - 7 )

For x = 4       y  = 16 - 23   y  = - 7

Point P is in the line

Point Q

For x = 1     y = - 4 *(1) - 23      ⇒   y = - 27

Then poin Q is not in the line

Equation 3

y - 1 = - 4 * ( x - 4 )

y  - 1 =  -4*x + 16    ⇒   y  = - 4*x + 17

Point P ( 4 , - 7 )

For x = 4

y = - 16 + 17   ⇒ y = 1    

Point P is not in the line

And Point Q ( 1 , 5 )

For  x = 1

y = - 4* ( 1 ) + 17    ⇒  y  = 13    Q is not in the line

3 0
3 years ago
On a trip to Griffith Observatory, Dave rode his bicycle six more than twice as many miles in the afternoon as in the morning. I
podryga [215]

Answer:

Dave rode his bicycle 17 miles in the morning

Dave rode his bicycle 40 miles in the afternoon

Step-by-step explanation:

* Lets study the information to solve the question

- Dave rod his bicycle in the morning and again in the afternoon

- Six more than twice as many miles in the afternoon as in the morning

  means the distance in the afternoon is 6 more than twice the distance

  in the morning

- The entire trip was 57 miles means the total distance in the morning

  and in the afternoon was 75

* To solve the question let the distance in the morning is x miles

∵ The distance in the morning = x miles

∵ The distance in the afternoon is 6 more than twice the distance

  in the morning

∴ The distance in the afternoon = 2x + 6 miles

∵ The distance in entire trip = 57 miles

- The distance in the morning and the distance in the afternoon = 57 miles

∴ x + (2x + 6) = 57 ⇒ simplify

∴ 3x + 6 = 57 ⇒ subtract 6 from both sides

∴ 3x = 51 ⇒ divide both sides by 3

∴ x = 17 miles

∵ The distance in the morning is x miles

∴ Dave rode his bicycle 17 miles in the morning

∵ The distance in the afternoon = 2x + 6 miles

- Substitute the value of x

∴ The distance in the afternoon = 2(17) + 6 = 34 + 6 = 40 miles

∴ Dave rode his bicycle 40 miles in the afternoon

4 0
3 years ago
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