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elena55 [62]
3 years ago
14

F(x)=-12/25(x-25)^2+32 in standard form

Mathematics
1 answer:
marin [14]3 years ago
4 0

Answer: y= -12x^2/25+24x-268

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Using the Completing the Square method, what are the zeros of the quadratic function f(x) = 2x2 + 8x - 3?
Ipatiy [6.2K]

Answer:

A.

Step-by-step explanation:

1) to rewrite the given equation:

f(x)=2(x²+4x)-3; ⇔ f(x)=2(x²+4x+4)-3-8; ⇔ f(x)=2(x²+4x+4)-11; ⇔ f(x)=2(x+2)²-11;

2) the roots are:

\left[\begin{array}{ccc}x=-\sqrt{\frac{11}{2}}-2 \\x=\sqrt{\frac{11}{2}}-2 \end{array}

3) finally, the correct answer is A.

4 0
2 years ago
Formula to find volume of a right triangular prism
ozzi
The formula is base x height
7 0
3 years ago
3.5+2=2x-10help me HELP ME JIGGABOOS
Genrish500 [490]
3.5 + 2 = 2x - 10
5.5 = 2x -10
+10 +10
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15.5 = 2x
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2 2

7.75 = x

7.75 is the answer
5 0
1 year ago
Domain and range of the function and it’s inverse
sineoko [7]

Answer:

B (second table in list)

Step-by-step explanation:

The function is a quadratic with vertex and y-intercept at (0,-2). The parabola opens up meaning only y - values above -2 are a part of the function. Since range is the set of all y values, then the range is y\geq -2.

The domain is not restricted and it is all real numbers.

The inverse of the function is where x and y are switched. So the range becomes the domain and vice verse.

This means the domain of the inverse is x\geq -2 and the range of the inverse is all real numbers.

8 0
3 years ago
If a = √3-√11 and b = 1 /a, then find a² - b²​
Serga [27]

If b=\frac1a, then by rationalizing the denominator we can rewrite

b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8

Now,

a^2 - b^2 = (a-b) (a+b)

and

a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8

a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8

\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}

5 0
2 years ago
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