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3 years ago

When is the difference of two decimals an integer?

2 answers:

8 0

The time the difference between two decimals is an integer is when the decimal value is the same for both numbers. For example, 1.20 - 0.20 = 1. 1.20 and 0.20 are decimals while 1 is an integer. 0.20 is the decimal value while 1 is the whole number.

Komok [63]3 years ago

5 0

If both numbers are themselves integers, then the difference will be an integer. Otherwise… If both real number have the same sign (both positive or both negative), the difference is an integer if every single digit after the decimal point is the same.

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Please please help!!!!!!

RideAnS [48]

Answer:

input = 5

Step-by-step explanation:

y = -2x - 6

When y = -16

-16 = -2x - 6

divide by -2 on both sides,

8 = x + 3

x = 8 - 3 = 5

6 0

2 years ago

Find the 14th term of the geometric sequence 7, 28, 112,

Fed [463]

Answer:

a(14)=469762048

Step-by-step explanation:

geometric sequence= a(n)=a*r^(n-1)

we have:

a=7

r=28/7=4

So:

a(14)=7*4^(14-1)

=7*4^(13)

=469762048

hope this helps

6 0

3 years ago

Multiple the following:

fredd [130]

Answer:

1.-9x^2

2.16x^4

3.x^2+4x

4.x^2+4x+4

5.x^2-x-12

6.x^4+2x^3-4x^2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Question 13 of 40

eduard

Answer:

b i think

Step-by-step explanation:

3 0

2 years ago

A school wishes to enclose its rectangular playground using 480 meters of fencing.

Harlamova29_29 [7]

Answer:

Part a) $A(x)=(-x^2+240x)\ m^2$

Part b) The side length x that give the maximum area is 120 meters

Part c) The maximum area is 14,400 square meters

Step-by-step explanation:

The picture of the question in the attached figure

Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x

we know that

The perimeter of the rectangular playground is given by

$P=2(L+W)$

we have

$P=480\ m\\L=x\ m$

substitute

$480=2(x+W)$

solve for W

$240=x+W\\W=(240-x)\ m$

<u><em>Find the area of the rectangular playground</em></u>

The area is given by

$A=LW$

we have

$L=x\ m\\W=(240-x)\ m$

substitute

$A=x(240-x)\\A=-x^2+240x$

Convert to function notation

$A(x)=(-x^2+240x)\ m^2$

Part b) What side length x gives the maximum area that the playground can have?

we have

$A(x)=-x^2+240x$

This function represent a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the length that give the maximum area that the playground can have

Convert the quadratic equation into vertex form

$A(x)=-x^2+240x$

Factor -1

$A(x)=-(x^2-240x)$

Complete the square

$A(x)=-(x^2-240x+120^2)+120^2$

$A(x)=-(x^2-240x+14,400)+14,400$

$A(x)=-(x-120)^2+14,400$

The vertex is the point (120,14,400)

therefore

The side length x that give the maximum area is 120 meters

Part c) What is the maximum area that the playground can have?

we know that

The y-coordinate of the vertex represent the maximum area

The vertex is the point (120,14,400) -----> see part b)

therefore

The maximum area is 14,400 square meters

Verify

$x=120\ m$

$W=(240-120)=120\ m$

The playground is a square

$A=120^2=14,400\ m^2$

8 0

3 years ago