All categories

3 years ago

Do the side lengths of 8, 15, and 20 forms a triangle?

Mathematics

2 answers:

liq [111]3 years ago

6 0

Answer:

yes

Step-by-step explanation:

when 1st and 2nd lengths addition is more than 3rd side

for example

6+8>10

trapecia [35]3 years ago

3 0

Answer:

Yes they do

Step-by-step explanation:

https://www.mathwarehouse.com/geometry/triangles/triangle-inequality-theorem-rule-explained.php?alenght=8&blength=15&clength=20#triangle-inequality-theorem

This demostration shows how.

You might be interested in

There were 24 green jelly beans which made up 8%of all the jelly beans in the bag

Zanzabum

There are 300 jelly beans in the bag.

24÷6=4
4×2=8
5×2=10
5×6=30
30×10=300

Proof:
24/300 = ?/100
24×100=2400
2400÷300=8

Hope this helps! :)

6 0

3 years ago

What two numbers multiply to get 60 and add to get 19?

marusya05 [52]

4 and 15. You can guess and check or use algebra.

4 0

3 years ago

Read 2 more answers

Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case

olga nikolaevna [1]

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> = Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

$=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0

3 years ago

Read 2 more answers

Can you help me please

xenn [34]

(2, -3)

The Answer is A

3 0

3 years ago

Two friends leave school at the same time Sarah is heading to do North and Beth is heading to West one hour later they are 5 mil

svet-max [94.6K]

5^-4^=25-16=9 root from 9 is 3 so it’s c) 3 miles

7 0

3 years ago