All categories

3 years ago

Sort the property that characterizes either a trapezoid or a kite.

Mathematics

1 answer:

Pie3 years ago

5 0

Trapezoid:

•Can have congruent diagonals. •Has one pair of opposite, parallel sides.

Kites:

•Has congruent adjacent sides.

•Has perpendicular diagonals.

You might be interested in

Given the function f(x)=x^2 and the transformed function g(x)=4x^2+1, explain how the parent function was transformed.

satela [25.4K]

Answer:

In the function g(x) = 4x² + 1, 4 means that all the points in the parent function were multiplied by 4. + 1 means that the parent graph will shift up by 1.

4 0

3 years ago

Please helppp me :( it’s c^4/c^3

ira [324]

Answer:

Step-by-step explanation:

Because $\frac{c}{c}$ is 1 the equation turns into...

$c^{4-3}$

which equals...

$c^{1}$

and $c^{1}$ is just c

7 0

2 years ago

The function g is defined as g(x)=5x² – 3. Find g(x+1).

maksim [4K]

Answer:

5x² + 10x + 2

Step-by-step explanation:

g(x + 1) = 5(x + 1)² - 3 = 5( x² + 2x + 1 ) - 3 = 5x² + 10x + 2

8 0

3 years ago

I WILL GIVE 175 POINTS TO ANYONE WHO HELPS ON ALL QUESTIONS

anzhelika [568]

Answer:

--No answers here, just an additional question---

Question:

What in the heck is this : 푦

8 0

3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago