4.

A probability experiment will consist of selecting one student at random from your math class. Is the probability that the selec

kumpel [21]

Not likely because in my class we have more boys then girls
m=27
f=5

8 0

3 years ago

Experimentally verify sum of three angles of an triangle is 180⁰.

svetlana [45]

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

4 0

3 years ago

You got paid twice this month one check was for $230.16 and the other was for $131.08 how much did you get paid all together

STALIN [3.7K]

Answer: The answer is 361.24

4 0

3 years ago

Read 2 more answers

40% is what number is 82?

Licemer1 [7]

32.8 because.40•82=32.8

8 0

4 years ago

Read 2 more answers

A rectangular field is 95 meters long and 65 meters wide. Give the length and width of another rectangular field that has the sa

Lena [83]

Answer:

There are a ton of answers. For instance, 1m by 159m.

Step-by-step explanation:

The formula for the area is $a=l*w$ , where $l$ and $w$ are constants.

The formula for perimeter is $p=2l+2w$

The question wants another rectangle that has the same perimeter, but a smaller area.

The perimeter of the two fields has to be:

$p=(2*95)+(2*65)\\p=190+130\\p=320$

Thankfully, the question never specified how much smaller the area has to be. So let's go to the logical extreme and make it 1 meter wide. To conserve the perimeter, we'd have to add the removed 64 meters to the original 95 meters of length, resulting in a rectangle with a width of 1m and a length of 159m. This results in an area of exactly 159 $m^{2}$ . The original rectangle had an area of 6175 $m^{2}$ . Mission accomplished!

But why stop there? Let's go even further down this rabbit hole!

Instead of 1, why don't we take it several steps further and use 0.0000001? At that point, the length would end up being 159.9999999m.

The area would be 0.00001599999999 $m^{2}$ .

Let's go further!!

Width: 0.00000000000000000001

Length: 159.99999999999999999999

Area: 0.00000000000000000001599999999999999999999 (approximate).

One more, for old time's sake.

Width: 0.00000000000000000000000000000000000000000000000001m

Length: 159.99999999999999999999999999999999999999999999999999m

Area: 0.00000000000000000000000000000000000000000000000015999999999999999999999999999999999999999999999999999 $m^{2}$

At that point, there's so little area in the plot of land that you could claim it has no area!

5 0

3 years ago