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Mrrafil [7]
3 years ago
10

If the sum of 4040 consecutive integer numbers is 2020, find out the absolute difference between the first number and the last n

umber.
Mathematics
1 answer:
bearhunter [10]3 years ago
4 0

Let x be the smallest of the consecutive integers summing to 2020. Then the largest integer in the sum is x+4039, so our answer is \boxed{4039}.

For completeness, though, let's solve for the value of x that satisfies the conditions of the problem. Recall that the sum of an arithmetic series with first term a_1, last term a_n, and n terms, is \frac{n(a_1+a_n)}{2}. Plugging in our known values gives us the equation \frac{4040(2x+4039)}{2}=2020, which simplifies to 2x+4039=1. Thus, x=-2019, so the 4040 consecutive integers are -2019,-2018,\dots,-1,0,1,\dots,2018,2019,2020. Notice how everything except the 2020 cancels out!

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pashok25 [27]

Answer:

First, take x off of both sides making it: “5 -4 = 3 + 10”

Then subtract 3, making it “3(-7) = 10”

Then do 3 multiply 3 by -7, getting -21 = 10

that is how you would find out the equation is false

6 0
3 years ago
The value of x varies directly with y, and when x= 1/3,y = 12. Write an equation to represent the direct variation.
shtirl [24]

Answer:

Equation: y = 36x

Step-by-step explanation:

The equation that represents a direct variation between variable x and variable y is given as y = kx. Where, k is the constant of proportionality.

Find the value of k given that x = ⅓ and y = 12.

Plug the values into the equation.

y = kx

12 = k\frac{1}{3}

Multiply both sides by 3

12*3 = k*1

36 = k

Substitute the value of k into y = kx

y = 36x

Equation: y = 36x

5 0
3 years ago
Which of these ARE NOT equivalent?<br>I NEED HELPPP
postnew [5]

Answer:

the first one I'm pretty sure

Step-by-step explanation:

I tried the other ones

3 0
3 years ago
Urgent please help me!!!!
enot [183]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
A circle is centered at the point O (0,0) and has a radius of the square root of 38?
nekit [7.7K]

Answer:

<em>The point (6,1) lies </em><em>inside</em><em> the circle.</em>

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle centered at the point (h,k) and with radius r, can be expressed with the equation:

(x-h)^2+(y-k)^2=r^2

The center of the given circle is (0,0) and the radius is r=\sqrt{38}, thus the equation of the circle is:

(x-0)^2+(y-0)^2=(\sqrt{38})^2

Simplifying:

x^2+y^2=38

Each point (x,y) meeting the condition of the equation lies exactly on its circumference. For example, the point (\sqrt{2},6) belongs to the circumference, but the origin (0,0) is inside the circle, and the point (5,5) is outside the circle.

We can tell the difference by evaluating the equation at each point. For the first point (\sqrt{2},6), the equation is:

(\sqrt{2})^2+6^2=2+36=38

The equation is satisfied.

For the origin (0,0):

0^2+0^2=0

For the point (5,5):

5^2+5^2=50

We can conclude that if the left side is greater than 38, the point is outside the circle, if it's less than 38, the point is inside the circle.

Let's test the point (6,1):

6^2+1^2=37

The point (6,1) lies inside the circle.

6 0
3 years ago
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