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maria [59]
3 years ago
5

The number square root of 3 goes on forever without repeating therefore it is rational?

Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0
Yes it is rational. b
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Four students had the following amounts of money in their pockets: $3.14, $.67, $2.45, and $1.14. How much money did they have t
Butoxors [25]
A. $7.40
Adding $3.14 to $0.67 equals $3.81
You then could add $2.45 to $3.81 and get $6.26
Finally you add $1.14 to $6.26 and get $7.40
7 0
3 years ago
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Please help me now ! Thank you
jarptica [38.1K]
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real. 

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result. 

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values


5 0
3 years ago
DEFINE A VARIABLE AND SET UP AND EQUATION TO SOLVE FOR THE FOLLOWING PROBLEM, DO NOT SOLVE!
statuscvo [17]

Answer:

5.8

Step-by-step explanation:

29/5=5.8

5 0
3 years ago
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The sum of the squares of two consecutive positive integers is 41. Find the two
Komok [63]

We have to present the number 41 as the sum of two squares of consecutive positive integers.

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25

16 + 25 = 41

<h3>Answer: 4 and 5</h3>

Other method:

n, n + 1 - two consecutive positive integers

The equation:

n² + (n + 1)² = 41     <em>use (a + b)² = a² + 2ab + b²</em>

n² + n² + 2(n)(1) + 1² = 41

2n² + 2n + 1 = 41     <em>subtract 41 from both sides</em>

2n² + 2n - 40 = 0     <em>divide both sides by 2</em>

n² + n - 20 = 0

n² + 5n - 4n - 20= 0

n(n + 5) - 4(n + 5) = 0

(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0

n = -5 < 0 ∨ n = 4 >0

n = 4

n + 1 = 4 + 1 = 5

<h3>Answer: 4 and 5.</h3>
7 0
2 years ago
Solve the following quadratic equation by rearranging and then factorising x^2 + 7x + 19 = 9
wolverine [178]

Answer:

your lucky you did have a online lessons earlll

Step-by-step explanation:

The following are the temperatures in °C for the first 14 days of January:

-6, -2.5, 2, 2.5, -0.5, 5, 10, -3, -7, 3, -1, 7, 1, 4.5

5 0
2 years ago
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