Can someone help me with this?

The graph of F(x) can be stretched vertically and flipped over the x axis to produce the graph of G(x) if F(x)=x^2 which of the

ladessa [460]

Answer:

g(x) = -5x²

(option B)

Step-by-step explanation:

we know that our original graph, f(x) = x² is a parabola.

So, we can consider what happens when we adjust the function/equation of a parabola.

when we "vertically stretch" a parabola, we are increasing the value of x.

think of it this way: the steepness of a slope is rise over run. If we rise ten, and run one, that's going be a lot more steep than if we rise 1, run 1.

Let's say our x = 5

if f(x)=x²

f(5) = 25

> y value / steepness is 25

f(x) = 3x²

f(5) = 75

> y value / steepness is 75

So, we are looking for an equation with an increase in x present.

When a parabola has been flipped over the x-axis, we know that the original equation now includes a negative

suppose that x = 1

if y = x² ; y = 1² = 1

if y = -(x²) ; y = -(1²) ; y = -1

So, when we set x to be negative, we make our y-values end up as negative also (which makes the graph look as if it has been flipped upside-down)

This means that we are looking for a function with a negative x value.

So, we are looking for a negative x-value that is multiplied by a number >1

The graph that fits our requirements is g(x) = -5x²

hope this helps!!

6 0

2 years ago

Is this statement is true -8< 8

prohojiy [21]

Yes 8>-8 because 8 is negative

6 0

4 years ago

roger sells 23 necklaces at a fair a total of $ 201.25. Each necklaces is sold for the same price. what is the price of one neck

MrMuchimi

Answer:

$8.75

Step-by-step explanation:

It is given that,

Roger sells 23 necklaces at a fair a total of $ 201.25.

We need to find the price of 1 necklace. We can use unitary method to find it.

23 necklaces = $201.25.

For cost of 1 necklace, divide $201.25 by 23 as follows :

$x=\dfrac{201.25}{23}\\\\=\$ 8.75$

So, the cost of 1 necklace is $8.75.

4 0

3 years ago

The larger leg of a right triangle is 2 feet longer than its smaller leg. The hypotenuse is 4 feet longer than the smaller leg.

maks197457 [2]

Hope it helps you:))

7 0

3 years ago

In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal. The same is true for a lower

SCORPION-xisa [38]

Answer:

Verified!

Step-by-step explanation:

Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.

Let's apply it for 2x2 matix:

$A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0$

So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)$ .

So, eigenvalues are $\lambda_1=a,\:\lambda_2=c$

Let's apply it for 3x3 matrix:

$A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0$

So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)(\lambda-f)$ .

So eigenvalues are $\lambda_1=a,\:\lambda_2=c,\:\lambda_3=f$

4 0

4 years ago