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3 years ago

What is −34, 0.5, 23,−73, 1.2 in order from least to greatest?

Mathematics

1 answer:

denpristay [2]3 years ago

5 0

Answer:

-73, -34, 0.5, 1.2, 23

Step-by-step explanation:

In Mathematics, the set of numbers are arranged in two ways. It means that the numbers can be arranged either in ascending order or descending order. If the numbers are arranged from the least to the greatest, then it is called ascending order. In this form, the numbers are in increasing order. The first number should be lesser than the second number.

If the numbers are arranged from the greatest to the least, then it is called descending order. In this form, the numbers are in decreasing form. The first number should be greater than the second number.

Also, read: Descending Order

Standard Form

The standard form to represent the least to the greatest arrangement of numbers is given by:

a < b < c < d < …..

Here,

a, b, c, d represent the numbers

Example: 2 < 5 < 7 < 8

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Julie ran 489 square feet how many square feet did Julie run rounded to the nearest 10 or 100

DerKrebs [107]

Answer:

490 or 500

Step-by-step explanation:

If we round to the 10th, Julie ran 490 feet.

I we round to the 100th, Julie ran 500 sq. ft.

5 0

3 years ago

W+3 /3 = -4 Solve what (W) = Im having trouble solving this can someone help Thanks :)

poizon [28]

Answer:

w = - 15

Step-by-step explanation:

Assuming you mean

$\frac{w+3}{3}$ = - 4 ( multiply both sides by 3 to clear the fraction )

w + 3 = - 12 ( subtract 3 from both sides )

w = - 15

5 0

2 years ago

Read 2 more answers

A particle moves with velocity vector

asambeis [7]

Answer:

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}$

Step-by-step explanation:

We are given that velocity vector of a particle

$\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}$

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle at time t.

We know that

Velocity = $\frac{Displacement }{time}=\frac{ds}{dt}$

Therefore, $\vec{v}=\frac{\vec{ds}}{dt}$

$\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt$

Integrate on both sides then we get

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C$

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Substitute the value of point at time t=0 then we get

C= $-\hat{j}+\hat{k}$

Substitute the value of C then we get

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}$

Therefore, the position of particle at time t

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}$

8 0

3 years ago

Find the slope of the line that passes through (42, -74) and (70, -80).

PolarNik [594]

- 3/14 is your final answer.

7 0

2 years ago

Pls help I'm stuck, I would like an explanation if possible!

fenix001 [56]

Answer:

Step-by-step explanation:

So they are showing you a chart in which the column y is determined by the column x, so you divide y/x in any of the rows so that you get the relation.

100/4=25 (which is the same for every row)

It works pretty much the same for the graph, but the relation is different.

Now instead of a chart you have to figure out the values of x and y by looking at the points that make up the line. For example we have a point that is in the coordinate (2, 60), and that represents that the x value is 2 and its correspondent y is 60. And to find the relation we do it as in the first step but now with 2 and 60.

60/2=30 (which is the same for every point)

Finally we just calculate the value of y both in the graph and the chart when x=12

So for the chart: 12*25= 300

And for the graph: 12*30= 360

The difference between those two is 60.

7 0

2 years ago