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katrin2010 [14]
3 years ago
6

What are two consecutive integers that have a sum of -67

Mathematics
1 answer:
valkas [14]3 years ago
5 0

Answer:

The two integers are -34,-33

Step-by-step explanation:

x 1st integer

x+1 2nd integer

The sum is -67

x+x+1 = -67

Combine like terms

2x+1 = -67

Subtract 1 from each side

2x+1-1 = -67-1

2x= -68

Divide by 2

2x/2 = -68/2

x = -34

x+1 = -34+1 = -33


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1/4 (one-fourth)

Convert 1/2 to 2/4
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What are the x- and y-intercepts of the line shown below?
natima [27]

Answer:

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Step-by-step explanation:

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2 years ago
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Find the product of (x + 5)(x - 5).<br> x2 - 10x + 25<br> x2 + 10x + 25<br> x2 - 25<br> x2 + 25
Mekhanik [1.2K]

Answer:

x²-25

Step-by-step explanation:

Hi there!

We are given (x+5)(x-5) and we want to find their product.

As these are binomials, we can multiply them using FOIL, which is First, Outer, Inner, Last

So let's start with the firsts, which would be (x+5)(x-5), or the 2 x's; x*x=x²

Next, let's do the outers, which would be (x+5)(x-5); x * -5 = -5x

Now the inners; (x+5)(x-5); 5 * x = 5x
Finally, the lasts; (x+5)(x-5); 5 * -5 = -25

Now add all of these terms together:

x²-5x+5x-25

We can combine the 5x and -5x together to become x²-25

So the answer is x²-25, which is a difference of squares expression (if we have (a-b)(a+b), then the result is a²-b²).

Hope this helps! :)

See more on the difference of squares here: brainly.com/question/16085851

8 0
2 years ago
Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2
fredd [130]

Answer:

Both get the same results that is,

\left[\begin{array}{ccc}140\\160\\200\end{array}\right]

Step-by-step explanation:

Given :

\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]

and initial population,

\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]

a) - After two times, we will find in each position.

P_2=[P].[M]^2=[P].[M].[M]

M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]

     =\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]

\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]

b) - With in migration process, 500 people are numbered. There will be after a long time,

After\;inifinite\;period=[M]^n.[P]

Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]

                                   =\left[\begin{array}{ccc}140\\160\\200\end{array}\right]

4 0
3 years ago
Bro why r people trying to online date on the app
frosja888 [35]

Answer:

wait really ? lol

Step-by-step explanation:

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