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rodikova [14]
3 years ago
14

HELP PLEASE 30 POINTS! WILL MARK BRAINLIEST IF CORRECT!!!!!

Mathematics
2 answers:
Flura [38]3 years ago
6 0

Answer:

this is a required answer.

KATRIN_1 [288]3 years ago
5 0

Answer:

See Below.

Step-by-step explanation:

We want to verify:

\cos(x-y)-\cos(x+y)=2\sin(x)\sin(y)

We will utilize the following identities:

\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)

And:

\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)

So, by substitution, we acquire:

(\cos(x)\cos(y)+\sin(x)\sin(y))-(\cos(x)\cos(y)-\sin(x)\sin(y))=2\sin(x)\sin(y)

Distribute:

\cos(x)\cos(y)+\sin(x)\sin(y)-\cos(x)\cos(y)+\sin(x)\sin(y)=2\sin(x)\sin(y)

The first and third term will cancel:

\sin(x)\sin(y)+\sin(x)\sin(y)=2\sin(x)\sin(y)

Combine like terms:

2\sin(x)\sin(y)\stackrel{\checkmark}{=}2\sin(x)\sin(y)

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A uniform beam of length L = 7.30m and weight = 4.45x10²N is carried by two ovorkers , Sam and Joe - Determine the force exert e
Mama L [17]

Answer:

Effort and distance = Load  x distance

7.30 x 4.45x10^2N = 3.2485 X 10^3N

We then know we can move 3 points to the right and show in regular notion.

= 3248.5

Divide by 2 = 3248.5/2 = 1624.25 force

Step-by-step explanation:

In the case of a Second Class Lever as attached diagram shows proof to formula below.

Load x distance d1 = Effort x distance (d1 + d2)

The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.

If the wheelbarrow is held steady (i.e. in Equilibrium) then the moment of the effort must be equal to the moment of the load :

Effort x its distance from wheel centre = Load x its distance from the wheel centre.

This general rule is expressed as clockwise moments = anti-clockwise moments (or CM = ACM)

 

This gives a way of calculating how much force a bridge support (or Reaction) has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!

The moment of the load on the beam (F) must be balanced by the moment of the Reaction at the support (R2) :

Therefore F x d = R2 x D

It can be seen that this is so if we imagine taking away the Reaction R2.

The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up.

The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example attached.

We know from our experience that a lighter person will have to sit closer to the end of the see-saw to balance a heavier person - or two people.

So if CM = ACM then F x d = R2 x D

from our kitchen scales example above 2kg x 0.5m = R2 x 1m

so R2 = 1m divided by 2kg x 0.5m

therefore R2 = 1kg - which is what the scales told us (note the units 'm' cancel out to leave 'kg')

 

But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated.

Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams (or bridges) have several loads acting upon them.

In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.e. F1=60kg, F2=40kg, d1=6m and d2=9m

 

CM = ACM

(F1 x d1) + (F2 x d2) = R2 x Length of beam

(60kg x 6m) + (40Kg x 9m) = R2 x 12m

(60kg x 6m) + (40Kg x 9m) / 12m = R2

360kgm + 360km / 12m = R2

720kgm / 12m = R2

60kg = R2 (note the unit 'm' for metres is cancelled out)

So if R2 = 60kg and the total load is 100kg (60kg + 40kg) then R1 = 40kg

4 0
2 years ago
Plzzzzz help it's an emergecy and plzzzzz show step by step
marshall27 [118]
So.. both sides divided by 4.
Now u have 14/68=x/6 -5
Multiply all by 6
Now u have 84/68=x-30
21/17+510/17=x
531/17=x
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7 0
2 years ago
It is math( help please
leva [86]

After 12 months, Quynh will have saved $800

Simply follow the pattern shown on the graph.

Month 1: 250

Month 2: 300

Month 3: 350

Month 4: 400

Month 5: 450

Month 6: 500

Month 7: 550

Month 8: 600

Month 9: 650

Month 10: 700

Month 11: 750

Month 12: 800

7 0
3 years ago
Read 2 more answers
one day in october, agent sanchez enters the code 3477 how do you know this is code is incorrect and and will not open the door
kozerog [31]
Well first off the cia will say that they didn’t send him, but we know the truth. Second off that number is the last 4 digits to the illuminatis president, Tom Hanks. Lastly, that number is also the number of kids Joe Biden has sniffed in the last year.
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2 years ago
Please help! Will mark brainliest!
Anestetic [448]

Each spinner has 5 numbers, the total combinations would be 5 x 5 = 25 combinations,

The answer is 25

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