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Trava [24]
2 years ago
14

Write any four rational numbers between ─5 and ─4.

Mathematics
2 answers:
Lera25 [3.4K]2 years ago
4 0

The rational numbers between -5 and -4 are as follows:

  • -4.2
  • - 4.8
  • - 4.4
  • - 4.6

<h3>What are rational numbers?</h3>

Rational numbers are numbers that can be expressed as the ratio of two integers.  These numbers are in the form of p/q, where p and q can be any integer and q ≠ 0.

Therefore, the rational numbers between -5 and -4 are as follows:

  • -4.2
  • - 4.8
  • - 4.4
  • - 4.6

Note the numbers are between -5 and -4.

learn more on rational numbers here: brainly.com/question/17450097

#SPJ2

Rasek [7]2 years ago
3 0

Answer:

Step-by-step explanation:

Write any four rational numbers between ─5 and ─4.

-4.1,

-4.2,

-4.3,

-4.4,

-4.5

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Can you please help me solve this and show the solution 4y⁴+6y?​
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Answer:

4y^4 + 6y = 256y + 6 = <u>262y</u>

Step-by-step explanation:

Remove the variables for a second and think of the problem as a normal expression; 4^4 + 6

Simplify the first term; 4^4 = 256

Now, add 6; 256 + 6 = 262

Add back in the variable; 262y

4 0
3 years ago
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A game has an expected value to you of ​$100. it costs ​$500 to​ play, but if you win you receive​ $100,000 (including your ​$50
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Let X be a random variable denoting the amount of winnings in any given round of the game. There is some probability p of winning where P(X=99,500)=p, while P(X=-500)=1-p, assuming these are the only two possible outcomes. So we have

\mathbb E[X]=\displaystyle\sum_xxP(X=x)=99,500p-500(1-p)=100

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3 years ago
Wendy has x pairs of shoes. Her friend Dylann has 6 fewer pairs of shoes. Which
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Step-by-step explanation:

6 0
3 years ago
Which lines are perpendicular to 7x-3y=18
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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
3 years ago
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