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Viktor [21]
2 years ago
6

Will the square of a negative number always be positive? Explain why.

Mathematics
2 answers:
qwelly [4]2 years ago
6 0

the square of a negative number always be positive

eg: -2² = -2×-2 = (-×- = +) +4

the cube of a negative number always be positive

eg: -2³ = -2×-2×-2 = (-×-×- = +) -8

hope it helps...!!!

ExtremeBDS [4]2 years ago
3 0

Answer:

<u><em>the square of a negative number always be positive</em></u>

<u><em>eg: -2² = -2×-2 = (-×- = +) +4</em></u>

<u><em>the cube of a negative number always be positive</em></u>

<u><em>eg: -2³ = -2×-2×-2 = (-×-×- = +) -8</em></u>

Step-by-step explanation:

<u><em>the Simple answer his right but mark me brailiest plssss</em></u>

<u><em>It will help my account alot</em></u>

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ludmilkaskok [199]
The answer is:  "3" .
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Use the Pythagorean theorem (for right triangles):

a² + b² = c² ;

in which "a = "side length 1" (unknown; for which we which to solve);

               "b" = "side length 2" = "√3" (given in the figure) ;

               "c" = "length of hypotenuse" = "2√3" (given in the figure);
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a² + b² = c² ; 

a² = c² − b² ;

Plug in the known values for "c" and "b" ;

a² = (2√3)²  −  (√3)² ;

Simplify:

(2√3)² = 2² * (√3)² = 2 * 2 * (√3√3)  = 4 * 3 =  12 .

 (√3)²  = (√3√3) = 3 .

a²  = 12 − 3 = 9 .

a² = 9

Take the "positive square root" of EACH SIDE of the equation; to isolate "a" on one side of the equation;  & to solve for "a" ; 

+√(a²)  = +√9 ;

     a = 3 . 
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The answer is:  "3" .
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3 0
3 years ago
I dont understand this, can i get some help please
Ray Of Light [21]

When we want to find the roots of a one-variable function, we look for where its graph intersects the x-axis. In this case, the graph intersects the x-axis at x=1.

The vertex of a parabola is the highest or lowest point on it, depending on whether the leading coefficient of the quadratic function is negative or positive. In this case, we see that the lowest point is (1,0).

For the y-intercept, just look for where the graph intersects the y-axis; in this case, that point is (0,1).

Using this information, the vertex-form equation of the parabola is y=1\cdot(x-1)^2+0, so the factors are two copies of x-1. In this case, the value of a in the equation y=a(x-h)^2+k was conveniently 1; if that's not the case, you'll want to plug in x=0 to solve for the value of a that gives the correct y-intercept.

Does that help clear things up?

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3 years ago
Find the slope from the graph shown.<br><br> A. -1<br> B. undefined<br> C. 1<br> D. 0
zimovet [89]

Answer:

it is underfined because if you look closely there is no slope in the graph

8 0
2 years ago
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Determine whether the pair of equations represent parallel lines, perpendicular lines, or neither.
Ivahew [28]

Answer:

Parallel

Step-by-step explanation:

Parallel lines have the same slope but different y-intercepts. If you multiply the top equation by 2, you get:

2(12x + 4y = 16)

24x + 8y = 32

This shows that both lines have the same slope, but then you find the y-intercepts, they are different:

1st equation y-int = 4

2nd equation y-int = 9/2 or 4.5

6 0
3 years ago
If x^2+1/x^2=3 find the value of x^2/(x^2+1)^2<br> Express answer as a common fraction.<br> Thanks!!
timama [110]

x^2+\dfrac1{x^2}=3\implies x^4+1=3x^2\implies x^4-3x^2+1=0

By the quadratic formula,

x^2=\dfrac{3\pm\sqrt5}2\implies x^2+1=\dfrac{5\pm\sqrt5}2

Then

(x^2+1)^2=\dfrac{25\pm10\sqrt5+5}4=\dfrac{15\pm5\sqrt5}2

\implies\dfrac{x^2}{(x^2+1)^2}=\dfrac{\frac{3\pm\sqrt5}2}{\frac{15\pm5\sqrt5}2}=\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}

Multiply numerator and denominator by the denominator's conjugate:

\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}\cdot\dfrac{15\mp5\sqrt5}{15\mp5\sqrt5}=\dfrac{45\pm15\sqrt5\mp15\sqrt5-25}{15^2-(5\sqrt5)^2}=\dfrac{20}{100}=\dfrac15

3 0
3 years ago
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