Answer:
Step-by-step explanation:
Solution by substitution method
3x-4y=8
and 18x-5y=10
Suppose,
3x-4y=8→(1)
and 18x-5y=10→(2)
Taking equation (1), we have
3x-4y=8
⇒3x=4y+8
⇒x=(
4y+8)/
3 →(3)
Putting x=
(4y+8
)/3 in equation (2), we get
18x-5y=10
18(
(4y+8)
/3) -5y=10
⇒24y+48-5y=10
⇒19y+48=10
⇒19y=10-48
⇒19y= -38
⇒y=-
38
/19
⇒y= -2→(4)
Now, Putting y=-2 in equation (3), we get
x=4y+8
x=
(4(-2)+8)
/3
⇒x=
(-8+8)/
3
⇒x=
0/
3
⇒x=0
∴x=0 and y= -2
371 is the answer you welcome
5)
a. The equation that describes the forces which act in the x-direction:
<span> Fx = 200 * cos 30 </span>
<span>
b. The equation which describes the forces which act in the y-direction: </span>
<span> Fy = 200 * sin 30 </span>
<span>c. The x and y components of the force of tension: </span>
<span> Tx = Fx = 200 * cos 30 </span>
<span> Ty = Fy = 200 * sin 30 </span>
d.<span>Since desk does not budge, </span><span>frictional force = Fx
= 200 * cos 30 </span>
<span> Normal force </span><span>= 50 * g - Fy
= 50 g - 200 * sin 30
</span>____________________________________________________________
6)<span> Let F_net = 0</span>
a. The equation that describes the forces which act in the x-direction:
(200N)cos(30) - F_s = 0
b. The equation that describes the forces which act in the y-direction:
F_N - (200N)sin(30) - mg = 0
c. The values of friction and normal forces will be:
Friction force= (200N)cos(30),
The Normal force is not 490N in either case...
Case 1 (pulling up)
F_N = mg - (200N)sin(30) = 50g - 100N = 390N
Case 2 (pushing down)
F_N = mg + (200N)sin(30) = 50g + 100N = 590N
Answer:
m(ABC) = 218°
Step-by-step explanation:
The measure of any arc with end points of an inscribed angle = 2 × inscribed angle
Inscribed angle = m<AMC = 109°
Intercepted arc = m(ABC)
Therefore,
m(ABC) = 2 × 109°
m(ABC) = 218°
(Interchange the x and y-values)
(Solve for y)
