All categories

3 years ago

How do you know when you will graph a parabola?

Mathematics

1 answer:

Alexeev081 [22]3 years ago

6 0

Answer:

Hope it helps!

Step-by-step explanation:

The graphs of quadratic functions are called parabolas. Here are some examples of parabolas. All parabolas are vaguely “U” shaped and they will have a highest or lowest point that is called the vertex. Parabolas may open up or down and may or may not have x -intercepts and they will always have a single y -intercept.

The graph of a quadratic function is a U-shaped curve called a parabola. The sign on the coefficient a of the quadratic function affects whether the graph opens up or down. If a<0 , the graph makes a frown (opens down) and if a>0 then the graph makes a smile (opens up).

Credits:

Graphs of Quadratic Functions | Boundless Algebra - Lumen Learning

Algebra - Parabolas - Pauls Online Math Notes

You might be interested in

The radius r of a sphere is 8.5mm. calculate the sphere's volume v

IRINA_888 [86]

Volume = 4/3 * PI * r^3
Volume = 4/3 * PI * 8.5^3
Volume = 4.18879 * 614.125 
Volume = 2,572.44 cubic centimeters

4 0

3 years ago

A car traveled a distance of 126 miles in three hours traveling at the same speed what distance will the car travel in seven hou

Charra [1.4K]

Answer:

the correct answer would be 294 because you would have to divide 126/3 to find out how many miles per hour then multiply it by 7 for 7 hours

hope this help

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If D is the midpoint of the segment AC and C is the midpoint of segment DB , what is the length of the segment AB , if AC = 3 cm

liraira [26]

AC= 3cm
DB= 3cm
so, AB= AC+DB = 3+3 = 6cm

4 0

3 years ago

Read 2 more answers

Please help due today!

otez555 [7]

The answer is 5.5 feet.

15.125= 1/2 * b* h

15.125= 1/2 * b* (5.5)

2(15.125)= 2 (1/2 * b *(5.5))

30.25= b * 11

30.25/11= b * 11/11

2.75 = b

You have to check it so when you plug the solved variable back in

15.125= 1/2 * (2.75) * (5.5)

15.125= 7.5625

So because the original b variable does not equal to the area of the triangle, you multiply the b variable by two to get 5.5 and when you plug that in, you get 15.125.

8 0

4 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

4 years ago