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3 years ago

Place decimal points in 431 and 205 so that the difference of the two numbers is 16.19

Mathematics

1 answer:

Gnoma [55]3 years ago

7 0

Answer:

Step-by-step explanation:

Note that 20 (an approximation for 205) less 4 (an approx. for 431) comes out to 16.

This tells us that the proper placement of the decimals is

20.50

- 4.31

----------

and this comes out to 16.19.

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You can show your work if u want to Find 2 numbers if Their sum is −7 and their difference is 14

Pepsi [2]

The larger number is (-7+14)/2 = 3.5.

The smaller number is (-7-14)/2 = -10.5.

_____

The two equations a+b=-7, a-b=14 can be solved to get these results. It is genrally convenient to solve them by adding one to the other, or subtracting one from the other.

7 0

2 years ago

Please solve the table correctly.

Hitman42 [59]

Answer:

See below.

Step-by-step explanation:

No. Triangles Perimeter

1 14

2 18

3 22

We see from the table, after the first perimeter (= 14), with each increase of 1 in the number of triangles, the perimeter increases by 4.

As an equation it is P = 14 + 4(n - 1) where P = perimeter and n = number of triangles.

4 0

3 years ago

Multiply polynomials

Burka [1]

Answer -2x^4 - 2x^3 + 22x^2 - 30x + 12

Here's your solution

=> (2x^2 - 4x + 2) - (x^2 + 3x - 6)

=> (2x^2 - 4x + 2) -x^2 - 3x + 6

=> firstly multiple whole with -x^2

=> -x^2(2x^2 - 4x + 2) = -2x^4 + 4x^3 - 2x^2

=> now with -3x

=> -3x(2x^2 - 4x + 2) = -6x^3 +12x^2 -6x

=> with 6

=> 6(2x^2 - 4x + 2) = 12x^2 - 24x + 12

=> now add all terms

=> -2x^4 - 2x^3 + 22x^2 - 30x + 12

hope it helps and your day will be full of happiness ^_^

4 0

2 years ago

Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4

djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

$\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty$