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3 years ago

9. A triangle is dilated using a scale factor of 4. The image is then dilated

Mathematics

1 answer:

dimulka [17.4K]3 years ago

8 0

Answer:

4-3=1

Step-by-step explanation:

it was dilated by 1

dunno what this was really askin tho

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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

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