1. Write the linear equation for a line with a slope of 1/2 that goes through the point (-1,<br> 0).

How can you use number patterns to find the least common factor

Aleks04 [339]

Answer:

The correct answer is: 360.

Explanation:

First we can express 120 as follows:

2 * 2 * 2 * 3 * 5 = 120

You can get the above multiples as follows:

120/2 = 60

60/2 =30

30/2 = 15

15/3 = 5 (Since 15 cannot be divisible by 2, so we move to the next number)

5/5 = 1

Take all the terms in the denominator for 120, you would get: 2 * 2 * 2 * 3 * 5 --- (1)

Second we can express 360 as follows:

360/2 = 180

180/2 = 90

90/2 =45

45/3 = 15 (Since 45 cannot be divisible by 2, so we move to the next number)

15/3 = 5

5/5 = 1

Take all the terms in the denominator for 360, you would get: 2 * 2 * 2 * 3 * 3 * 5 --- (2)

Now in (1) and (2) consider the common terms once and multiple that with the remaining:

2*2*2*3*5 = Common between the two

3 = Remaining

Hence (2*2*2*3*5) * (3) = 360 = LCM (answer)

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

If 2 pizzas cost $16.22 how much would one cost

koban [17]

16.22 divided by 2 = your answer. (8.11)

6 0

3 years ago

Read 2 more answers

(3x/2)^4 any answers

Anika [276]

$\bf \left( \cfrac{3x}{2} \right)^4\implies \left( \cfrac{(3x)^4}{2^4} \right)\implies \left( \cfrac{3^4x^4}{2^4} \right)\implies \cfrac{81x^4}{16}$

6 0

3 years ago

Read 2 more answers

How do you write in slope intercept with only the rate of change and one point. Like rate of change is -5 and the point is (3,0)

Simora [160]

Answer:

A linear equation in the slope-intercept form is written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

Now, we know that the rate of change (the slope) is -5

Then we just replace a by -5

y = -5*x + b

Now we also know that this line passes through a point, and the point is (3, 0)

This means that the point (3, 0) is a solution for the line equation, so when x = 3, we also have y = 0.

Replacing these values in our equation we get:

0 = -5*3 + b

0 = -15 + b

15 = b

Now we know the value of b, so we can replace it in the line equation to get:

y = -5*a + 15

Which is the complete equation of the line.

5 0

3 years ago

1. Write the linear equation for a line with a slope of 1/2 that goes through the point (-1, 0).