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azamat
2 years ago
7

Find the value of "x" in x² -2 = 47Thanks :)​

Mathematics
1 answer:
Dennis_Churaev [7]2 years ago
7 0

~~~~~x^2 -2 = 47\\\\\implies  x^2 = 47 +2 \\\\\implies x^2 = 49\\\\\implies x = \pm\sqrt{49}\\\\\implies x = \pm 7\\\\\text{Hence, }~ x = 7,~~ x = -7

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Item 17 The amount of chlorine in a swimming pool varies directly with the volume of water. The pool has 2.5 milligrams of chlor
zvonat [6]
8000 gal *\frac{ 1 L}{0.264 gal}*\frac{2.5 mg}{1 L} = 75,757 mg

Rounding to nearest thousand gives 76,000 mg of chlorine
7 0
3 years ago
Tell whether the lines of the pair of equations are parallel, perpendicular, or neither.
tino4ka555 [31]
First Equation:

y= -\frac{3}{4} x+2

Second Equation can be written as:

y= \frac{3}{4}x+2

Slope of first equation is -3/4 and slope of second equation is 3/4. 
Slope of parallel lines must be equal, and slope of perpendicular lines are the negative reciprocal of each other. None of these conditions can be seen for given two equations.

So, the two lines are neither parallel nor perpendicular.
So correct option is C
7 0
4 years ago
Read 2 more answers
What is the measure, in degrees, of ∠x in the figure?
gizmo_the_mogwai [7]

Answer:

I believe the answer is 127 degrees.  

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
solve image :) What is the equation of the line that has a slope of 3 and passes through the point (-4, 1)?
Lesechka [4]

Answer:

<em>C.  y = 3x + 13.</em>

Step-by-step explanation:

The point slope form of a line is

y - y1 = m(x - x1)  where m = the slope,  and (x1, y1) is a point on the line.

So, substituting:

y - 1 = 3(x - -4)

y - 1 = 3(x + 4)

y = 3x + 12 + 1

y = 3x + 13.

6 0
3 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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