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monitta
2 years ago
8

What is the vertex of g(x)=8x^2-64x

Mathematics
2 answers:
Mama L [17]2 years ago
8 0
<span>The (x,y) point of the vertex is usually labelled as:
Vertex = (h, k)
h = -b /  2a
h = --64 / 2*8
h = 64 / 16
</span><span>h = 4

</span>
<span>To get “k”, input the “h” value into the equation.
<span>k = 8*(4^2) -64*4
</span> <span>k = 8*(4^2) -64*4
</span><span>k = 128 -256
</span><span>k = -128
</span></span><span>Vertex = (4, -128)

</span>
Shkiper50 [21]2 years ago
5 0

Answer:

Vertex = (4, -128)

Step-by-step explanation:

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What is the radius of a circle given by equation x^2 + y^2 - 2x + 8y - 47 = 0
Lera25 [3.4K]
X^2 + y^2 - 2x + 8y - 47 = 0
x^2 + y^2 - 2x + 8y = 47
(x^2 - 2x) + (y^2 + 8y) = 47
(x^2 - 2(1)x) + (y^2 + 2(4)y) = 47
(x^2 - 2(1)x + 1^2) + (y^2 + 2(4)y + 4^2) = 47 + 1^2 + 4^2
(x - 1)^2 + (y + 4)^2 = 64 = 8^2
r=8
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2 years ago
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Pls help asap math isn't my thing
stira [4]

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3 years ago
Si -x+y+z=75, entonces -(x+3)+(y-5)+(z-6)=
jenyasd209 [6]

Answer:

67

Step-by-step explanation:

● -(x + 3) + (y - 5) + (z - 6)

● -x - 3 + y - 5 + z - 6

● (-x + y + z) + (3 -5 -6)

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2 years ago
Solve for x in the equation x2 - 4x-9 = 29.
erastova [34]

Answer:

x=2+\sqrt{21}\\\\x=2-\sqrt{21}

Step-by-step explanation:

One is given the following equation;

x^2-4x-9=29

The problem asks one to find the roots of the equation. The roots of a quadratic equation are the (x-coordinate) of the points where the graph of the equation intersects the x-axis. In essence, the zeros of the equation, these values can be found using the quadratic formula. In order to do this, one has to ensure that one side of the equation is solved for (0) and in standard form. This can be done with inverse operations;

x^2-4x-9=29

x^2-4x-38=0

This equation is now in standard form. The standard form of a quadratic equation complies with the following format;

ax^2+bx+c

The quadratic formula uses the coefficients of the quadratic equation to find the zeros this equation is as follows,

\frac{-b(+-)\sqrt{b^2-4ac}}{2a}

Substitute the coefficients of the given equation in and solve for the roots;

\frac{-(-4)(+-)\sqrt{(-4)^2-4(1)(-38)}}{2(1)}

Simplify,

\frac{-(-4)(+-)\sqrt{(-4)^2-4(1)(-38)}}{2(1)}\\\\=\frac{4(+-)\sqrt{16+152}}{2}\\\\=\frac{4(+-)\sqrt{168}}{2}\\\\=\frac{4(+-)2\sqrt{21}}{2}\\\\=2(+-)\sqrt{21}

Therefore, the following statement can be made;

x=2+\sqrt{21}\\\\x=2-\sqrt{21}

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