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erik [133]
3 years ago
13

Identify two laws created as part of the intolerable acts

Mathematics
1 answer:
mixer [17]3 years ago
3 0
This is history but here you go:

-closed Boston harbor
-banned town meetings
-people forced to quarter soldiers in homes
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The sun of 2 positive numbers is 17. Their product is 60
butalik [34]
Answer is 5 and 12 thank you
5 0
2 years ago
PLZ HELP ASAP!!!!!!!! WILL MARK YOU BRAINLIEST THX!
anygoal [31]

Answer:

The first question is the third choice. Divide by two to get proper squaring. Anytime there is a number in front of the x in a situation like that, always divide by that number on both sides first. Then go from there.

The second question is also the third choice. The square root of 64 is 8. It can't be negative because square rooting is always positive unless you want a weird "i" in there. That's for algebra 2 and beyond so don't worry about it mate.

7 0
3 years ago
What is the product of 4/7 and 1/6?
jek_recluse [69]

Answer: 0.09523

Step-by-step explanation:

5 0
3 years ago
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
Drag the tiles to the correct boxes to complete the pairs.
erastova [34]

Answer:

√8 ==> 2 units, 2 units

√7 ==> √5 units, √2 units

√5 ==> 1 unit, 2 units

3 ==> >2 units, √5 units

Step-by-step explanation:

To determine which pair of legs that matches a hypotenuse length to create a right triangle, recall the Pythagorean theorem, which holds that, for a right angle triangle, the square of the hypotenuse (c²) = the sum of the square of each leg length (a² + b²)

Using c² = a² + b², let's find the hypotenuse length for each given pairs of leg.

=>√5 units, √2 units

c² = (√5)² + (√2)²

c² = 5 + 2 = 7

c = √7

The hypothenuse length that matches √5 units, √2 units is √7

=>√3 units, 4 units

c² = (√3)² + (4)²

c² = 3 + 16 = 19

c = √19

This given pair of legs doesn't match any given hypotenuse length

=>2 units, √5 units

c² = (2)² + (√5)²

c² = 4 + 5 = 9

c = √9 = 3

legs 2 units, and √5 units matche hypotenuse length of 3

=>2 units, 2 units

c² = 2² + 2² = 4 + 4

c² = 8

c = √8

Legs 2 units, and 2 units matche hypotenuse length of √8

=> 1 unit, 2 units

c² = 1² + 2² = 1 + 4

c² = 5

c = √5

Leg lengths, 1 unit and 2 units match the hypotenuse length, √5

3 0
3 years ago
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