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3 years ago

Help pls itS BEEN 2 HOURS LUV U <3

Mathematics

1 answer:

nataly862011 [7]3 years ago

7 0

Step-by-step explanation:

Hey there!

The equation of a circle which passes through any point having centre (h,k) is;

$( {x - h)}^{2} + ( {y - k)}^{2} = {r}^{2}$

In first blank ( x-_ ) keep "-2". In second blank (y - _ ) keep "4" and in last blank ( _ ) keep "36".

Hope it helps..

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The minimum value of Y in the equation y = x2 - 6x + 8

murzikaleks [220]

Given

$y=x^2-6x+8$

To obtain the minimum value of y, we first take the derivative of y

The derivative of y is:

$y^{\prime}=2x-6$

Equating

$y^{\prime}\text{ = 0}$

gives the minimum value we require.

Doing that, we have:

$2x-6=0$

So that

$\begin{gathered} 2x=6 \\ x=\frac{6}{2} \\ =3 \end{gathered}$

Therefore, the minimum value is x = 3

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1 year ago

If the distance covered by an object in time t is given by s(t)=t^2+5t , where s(t) is in meters and t is in seconds, what is th

alisha [4.7K]

Answer:

E. 44 meters

Step-by-step explanation:

The function that models the distance covered by the object is

$s(t)=t^2+5t$

where s(t) is in meters and t is in seconds.

The distance covered by the object after 1 second is

$s(1)=1^2+5(1)=6m$

The distance covered by the object after 1 second is

$s(1)=5^2+5(5)=50m$

The distance covered between 1 second and 5 seconds is

50-6=44m

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3 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .

$15x(x^2-3x+2x-6)=0$

$15x((x^2-3x)+(2x-6))=0$

$15x(x(x-3)+2(x-3))=0$

$15x(x-3)(x+2)=0$

Now, we will use zero product property to find the zeros of our given function.

$15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0$

$15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0$

$\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0$

$x=0\text{ (or) }x=3\text{ (or) }x=-2$

Therefore, the zeros of our given function are $x=-2,0,3$ .

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3 years ago

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The sampling distribution standard deviation is the population standard deviation divided by the root of the sample size.

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3 years ago

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