URGENT plz help me will give brainlist

DochEvi [55]

Answer:

well the shape is acute so it will be quite low work out the opposite angles and you will find out that the lines are parallels there for meaning the answer is the lowest angle

Step-by-step explanation:

3 0

2 years ago

Determine if the expression is a polynomial or not. If it is a polynomial, state the

katrin2010 [14]

Answer:

Step-by-step explanation:

yes that

3 0

2 years ago

Any time a scientific theory is challenged it means it's not a good theory. Why?

likoan [24]

If a theory is challenged, then that means that the theory may have a specific flaw, making it "not good"

5 0

2 years ago

Read 2 more answers

Suppose that the functions h and f are defined as follows. h(x) =3/x, x notequalto 0 f(x) = x^2 - 1 Find the compositions h comp

Klio2033 [76]

Answer:

hf(x) = 3/(x² - 1)

fh(x) = (9 - x²)/x²

Step-by-step explanation:

h(x) = 3/x

f(x) = x² - 1

h(f(x)) = 3/f(x)

hf(x) = 3/(x² - 1)

f(h(x)) = (h(x))² - 1

= (3/x)² - 1

= 9/x² - 1

fh(x) = (9 - x²)/x²

4 0

2 years ago

An ellipse has foci located at (2,4) and (2,16) and vertices located at (2,0) and (2,20) this ellipse is vertical

pochemuha

Answer:

The equation of the ellipse is $\frac{(x-2)^{2}}{8^{2}}+\frac{(y-10)^{2}}{10^{2}} = 1$ .

Step-by-step explanation:

The statement is incomplete, the most probable outcome may be the equation of the ellipse based on information given.

The equation of an ellipse whose major axis is vertical and is centered at a point different from origin is defined by:

$\frac{(x-h)^{2}}{b^{2}} + \frac{(y-h)^{2}}{a^{2}} = 1$ (1)

Where:

$(h,k)$ - Coordinates of the center of the ellipse.

$(x,y)$ - Coordinates of a point in the line of the ellipse.

$a$ - Length of the major semiaxis.

$b$ - Length of the minor semiaxis.

The coordinates of the center of the ellipse is midpoint of the segment between vertices, which are collinear with foci:

$(h,k) = \frac{1}{2}\cdot V_{1} (x,y) + \frac{1}{2}\cdot V_{2}(x,y)$ (1)

Where $V_{1} (x,y)$ and $V_{2} (x,y)$ are the coordinates of the vertices.

If we know that $V_{1} (x,y) = (2,0)$ and $V_{2} (x,y) = (2, 20)$ , then the coordinates of the center ellipse are:

$(h,k) = (2, 10)$

The length of the semimajor axis can be determined by using the following vectorial expression, which is equivalent to the Pythagorean Theorem:

$a = \sqrt{[(h,k)-V_{1}(x,y)]\,\bullet\,[(h,k)-V_{1}(x,y)]}$ (2)

If we know that $(h,k) = (2, 10)$ and $V_{1} (x,y) = (2,0)$ , then the length of the semimajor axis is:

$a = \sqrt{(2-2)^{2}+(10-0)^{2}}$

$a = 10$

And the length of the minor semiaxis is found by means of this Pythagorean identity:

$b = \sqrt{a^{2}-c^{2}}$ (3)

Where $c$ is the length between the center and any of the foci. This distance can be found by using this vectorial formula:

$c = \sqrt{[(h,k)-F_{1}(x,y)]\,\bullet\,[(h,k)-F_{1}(x,y)]}$ (4)

If we know that $(h,k) = (2, 10)$ and $F_{1} (x,y) = (2,4)$ , then the length between the center and any of the foci is:

$c = \sqrt{(2-2)^{2}+(10-4)^{2}}$

$c = 6$

And the length of the minor semiaxis is:

$b = \sqrt{10^{2}-6^{2}}$

$b = 8$

FInally, the equation of the ellipse is $\frac{(x-2)^{2}}{8^{2}}+\frac{(y-10)^{2}}{10^{2}} = 1$ .

6 0

2 years ago