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3 years ago

What polygon is formed?

Mathematics

1 answer:

Nuetrik [128]3 years ago

8 0

Answer:

Polygons are figures that are formed by three or more line segments. These line segments, even called sides, meet at their endpoints to form vertices. If we compare the number of sides, diagonals and triangles of a quadrilateral a pentagon and a hexagon we can find a system between the different polygons.

Step-by-step explanation:

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Put commas and write the following numbers in Indian system of

Alenkinab [10]

1. Indian System:- 

(a) 23,45,678

(b) 56,78,090

2. International System:- 

(a) 234,589

(b) 9,807,062

8 0

3 years ago

Jimmy threw a baseball in the air from the roof of his house. The path followed by the baseball can be modeled by the function f

erastovalidia [21]

Answer:

Step-by-step explanation:

The first part of A is easy. Look at the quadratic function, and the constant, the very last number with no t stuck to it represents the height from which the object in question was originally launched. Our constant is 40, so the height of the roof from which the baseball was thrown is 40 feet. Part 2 of A is not quite as simple because it requires factoring using the quadratic formula.Before we do that, let's make our numbers a bit more manageable, shall we? Let's factor out a -8 to get

$f(t) = -(t^2-6t-5)$ and a = 1, b = -6, c = -5.

Filling in the quadratic formula now looks like this:

$t=\frac{6+-\sqrt{6^2-4(1)(-5)} }{2(1)}$ and

$t=\frac{6+-\sqrt{36+40} }{2}$ and

$t=\frac{6+-\sqrt{56} }{2}$ so the 2 solutions are

$t=\frac{6+\sqrt{56} }{2}=6.74sec$ and

$t=\frac{6-\sqrt{56} }{2}=-.742sec$ and since we know time can NEVER be negative, the time it takes for the baseball to hit the ground from a height of 40 feet is 6.74 seconds. Onto part B.

In order to determine exactly how high the baseball did go, we have to find the vertex of the function. We do this by completing the square and getting the function into vertex, or work, form. Begin by setting the quadratic equal to 0, moving over the constant, and then factoring out the leading coefficient. The rule for completing the square are kinda picky in that you have to have a 1 as the leading coefficient, and righ now ours is a -8. So following the rules I stated above:

$-8(t^2-6t)=-40$ Next is the take half the linear term, square it, and then add it to both sides. Our linear term is a -6. Half of -6 is -3, and -3 squared is 9, so we add 9 into the parenthesis first:

$-8(t^2-6t+9)=-40+??$

Because this is an equation, we can't add 9 to one side without adding the equivalent to the other side. But, we cannot forget about that -8 sitting out front there, refusing to be ignored. We didn't just add in a 9, we actually added in a -8 times 9 which is -72. That's what goes on the right side in place of the ??.

$-8(t^2-6t+9)=-40-72$

The reason we complete the square is found on the left side of the equals sign. We have, in the process of completing the square, formed a perfect square binomial that will serve as the h in our vertex (h, k) where h is the number of seconds it takes for the baseball to reach its max height of k, whatever k is. That's what we have to find out. Putting the left side into its simplified perfect square binomial and adding the numbers on the right gives us:

$-8(t-3)^2=-112$

For the last step, add over the -112 and set it back equal to f(t):

$-8(t-3)^2+112=f(t)$ From that we determine that the vertex is (3, 112). The max height of this baseball was 112 feet...so no, it did not make it up to the height of 120 feet that Jimmy wanted for the baseball.

6 0

2 years ago

There are 500 sheet of paper in the pack Hannah bought. She has used 137 sheets already. How many sheets of paper does Hannah ha

Grace [21]

Just work out 500 takeaway 137

5 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Debora [2.8K]

Yo sup??

you can solve this quesion by having the knowledge of trigonometric ratios

let angle B be x, then

tanx=7/3

x=66.8

Hope this helps

4 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago