All categories

3 years ago

A quadrilateral with two pairs of parallel sides and no right angles

Mathematics

1 answer:

kolezko [41]3 years ago

4 0

A Parallelogram. 2 obtuse angles, 2 acute angles. 2 pairs of parallel sides

You might be interested in

how do I solve 7v-(2v-16)=5(v+4) please explain if it is a one solution, no solution or infinite solution

Sergio039 [100]

Do the indicated mult. first:

7v - 2v + 16 = 5v + 20
5v + = 5v +20. This is never true. NO solution.

3 0

2 years ago

Find the value of x in each case

Juli2301 [7.4K]

Answer:

Pag x na wag na balikan wag ka marupok siz

6 0

2 years ago

PLEASE HELP (100 POINTS)

Artyom0805 [142]

$\\ \rm\Rrightarrow f(x)=-5cos\dfrac{x}{2}+5$

Compare to standard equation of SHM

$\\ \rm\Rrightarrow y=Acos(\omega x+\phi)$

Here

Amplitude=A=-5

$\\ \rm\Rrightarrow \omega x=\dfrac{x}{2}$

$\\ \rm\Rrightarrow \omega=1/2$

Well the above one are for extra knowledge .

Solution:-

$\\ \rm\Rrightarrow 10=-5cos\dfrac{x}{2}+5$

$\\ \rm\Rrightarrow 10-5=-5cos\dfrac{x}{2}$

$\\ \rm\Rrightarrow 5=-5cos\dfrac{x}{2}$

$\\ \rm\Rrightarrow cos\dfrac{x}{2}=-1$

$\\ \rm\Rrightarrow \dfrac{x}{2}=cos^{-1}(-1)$

$\\ \rm\Rrightarrow \dfrac{x}{2}=\pi$

$\\ \rm\Rrightarrow x=2\pi$

So.

$\\ \rm\Rrightarrow x=2\pi+2\pi n$

3 0

2 years ago

Read 2 more answers

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D.

yKpoI14uk [10]

The correct statement is:

There are four ways to choose the committee.
There are three ways to form the committee if person D must be on it.
If persons B and C must be on the committee, there are two ways to form the committee.

It is given that:

A membership committee of three is formed from four eligible members.

1) There are four ways to choose the committee.

This statement is true.

Since we have to choose 3 members out of the 4 members so we can use the method of combination.

2) There are three ways to form the committee if person D must be on it.

This statement is also true.

Since D has to be in the committee, this means we have to choose 2 more people out of the three people to form the committee.

3) If seven members are eligible next year, then there will be fewer combinations.

This statement is wrong.

Since we have to choose 3 members out of 7 members so the number of possible combinations will be:

There are 35 combinations possible.

4) If persons B and C must be on the committee, there are two ways to form the committee.

If B and C have to be in the committee then we have to choose just one person out of the two people left.

Hence, the statement is true.

5) If persons A and C must be on the committee, then there is only one way to form the committee.

If A and C have to be on the committee then as in the last option we have to choose any one of the two-person left.

So possible number of ways is 2.

Hence, the statement is false.

Learn more about committee here: brainly.com/question/425830

#SPJ4

<em>Your question is incomplete. Please read below to find the missing content.</em>

<em />

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D. The possible outcomes include S = {ABC, ABD, ACD, BCD}.

Which statements about the situation are true? Check all that apply.

There are four ways to choose the committee.

There are three ways to form the committee if person D must be on it.

If seven members are eligible next year, then there will be fewer combinations.

If persons B and C must be on the committee, there are two ways to form the committee.

If persons A and C must be on the committee, then there is only one way to form the committee.

8 0

1 year ago

Sqrt 6^2 + 8^2 jdjdjsns

slavikrds [6]

Answer:

Step-by-step explanation:

100

7 0

1 year ago