All categories

3 years ago

Order the quadratic functions from least to greatest based on the number of x-intercepts of each function.

Mathematics

1 answer:

MAVERICK [17]3 years ago

4 0

Y-intercept ... vertex ... g(t) ...

You might be interested in

In their latest play, the Giants lost another 5 yards. This is their third time losing 5 yards. They also gained 20 yards. Give

Vedmedyk [2.9K]

ANSWER: +5

EXPLANATION:

(-5) + (-5) + (-5) = (-15)

+20 + (-15) = +5

4 0

3 years ago

Find the difference.

Scorpion4ik [409]

Answer:

Is that the letter (x) or Is that times?

If it is times; then the difference is:

Option "C" -12x2-x-7

Step-by-step explanation:

why, because it is not solvable

3 0

3 years ago

I need help ASAP will reward 30 points <br> Question attached

mash [69]

Answer:

AB=24cm and BC=6cm

Step-by-step explanation:

6 0

4 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$