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2 years ago

Which of the following functions best describes this graph?

Mathematics

1 answer:

Katarina [22]2 years ago

4 0

We see that the zeros of the function are 4 and 5.

Now lets us write that into the equation

$y=(x-4)(x-5)$

Now multiply that out

$y = x^2-5x-4x + 20 = x^2-9x+20$

Hope that helped!

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Softa [21]

Answer:

Step-by-step explanation:

option 4

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3 years ago

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diamong [38]

Answer:

$y=\frac{6}{5}x-8$

Step-by-step explanation:

we know the slope (or m in y=mx+b) is 6/5 we just need to solve for b

we will do this by plugging in the required coordinates

-2=(6/5)*5+b

-2=6+b

-8=b

put it all together

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3 years ago

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Find three consecutive even integers such that the square of the third is 60 more that the square of the second

lana [24]

Answer:

-4,4,16

Step-by-step explanation:

They are all even integers.

-4^2=16

4^2=16

16^2=256

the square of the third,16 is 256 which is more than the square of the second,4=16

4 0

4 years ago

Simplify 25^ - 1/2 Please explain! I know the answer is 1/5 but why?

lisov135 [29]

$(25)^{-\frac{1}{2} }$

When the exponent is negative, you have to make the number a fraction(1 over the number).

For example:

$25^{-2} = \frac{1}{25^{2} }$

$x^{-1} =\frac{1}{x^{1} }$

This gets rid of the negative, but it leaves the exponent

$(25)^{-\frac{1}{2}} = \frac{1}{25^{1/2}}$

When an exponent is a fraction, it is the square root.

For example:

$x^{\frac{1}{3}} = \sqrt[3]{x}$

$x^{\frac{1}{2}} =\sqrt{x}$

$x^{\frac{1}{4}}=\sqrt[4]{x}$

So...

$\frac{1}{25^{1/2}} = \frac{1}{\sqrt{25} } = \frac{1}{5}$

3 0

3 years ago

The slope of the normal curve y=3x^2-5x is at the point (0,0)

Rufina [12.5K]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Consider the curve $y=3x^2-5x$

Differentiate with respect to $x$

$\frac{dy}{dx} =3(2x)-5=6x-5$

( Use $\frac{d}{dx}(x^n)=nx^{n-1}$ )

Slope of normal curve is given by $\frac{-1}{\frac{dy}{dx} }$

Slope of normal curve = $\frac{-1}{6x-5}$

To find slope of normal curve at point $(0,0)$ ,put $x=0$ in $\frac{-1}{6x-5}$

Slope of normal curve at point $(0,0)$ = $\frac{-1}{6(0)-5} =\frac{-1}{-5}=\frac{1}{5}$

8 0

3 years ago