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3 years ago

ZA and ZB are complementary angles. If m_A = (62 – 8)° and m

Mathematics

1 answer:

tino4ka555 [31]3 years ago

7 0

Step-by-step explanation:

A=62-8=54

if the angles are complementary that means that their sum is 90degrees, so 54+5x+21=90

5x=90-75=15

=>B=15+21=36

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3 years ago

Using approciate properties find -4/7*3/5+4/7*-1/5+2/5*-4/5

slega [8]

Answer: 0.548

Step-by-step explanation:

First you need to remember the formula for PEMDAS(Parenthesis,exponents,multiplication,divison,addition,subtractio)

you can do either multiplication or divison whichever comes first, same with addition+subtraction.

the underlined means to simplify

1.-0.57x3/5+4/7x-1/5+2/5x-4/5

2.-1.71/5+4/7x-1/5+2/5x-4/5

3.-0.342+4/7x-1/5+2/5x-4/5

4.-0.342+0.57x-1/5+2/5x-4/5

5.-0.342+0.57x-0.2+2/5x-4/5

6.-0.342+0.114+2/5x-4/5

7.-0.342+0.114+0.4x-4/5

8.-0.342+0.114+-1.6/5

9.-0.342+0.114+-0.32

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i did this all in one go so......

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3 years ago

SOLVED

Nastasia [14]

5)
a. The equation that describes the forces which act in the x-direction:
 Fx = 200 * cos 30 

b. The equation which describes the forces which act in the y-direction: 
 Fy = 200 * sin 30 

c. The x and y components of the force of tension: 
 Tx = Fx = 200 * cos 30 
 Ty = Fy = 200 * sin 30 

d.Since desk does not budge, frictional force = Fx
= 200 * cos 30 

 Normal force = 50 * g - Fy
= 50 g - 200 * sin 30
____________________________________________________________
6) Let F_net = 0
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

4 0

3 years ago