Can you form a triangle with these given side lengths? 12 , 8 , 16 Please help me!!!!

Mathematics

2 answers:

Scilla [17]4 years ago

6 0

Answer:

yes

Step-by-step explanation:

12+8>16

the two short sides have to be greater than the longest dude when combined and that is true so you can form a triangle

professor190 [17]4 years ago

3 0

Answer:

Yes it can

Step-by-step explanation:

You might be interested in

The data in the table can be entered into a calculator to determine a linear equation of best fit where x represents the number

Yakvenalex [24]

Based on the data given, the conclusion that can be drawn from the correlation coefficient associated with the linear equation is C. There's a weak negative correlation between the variables.

<h3>How to illustrate the information?</h3>

From the information given, it can be seen that the value of y is increasing with the increase in the value of x.

This implies that the correlation is positive. Also, the change in the values of y with x are scattered a lot.

Therefore, the conclusion that can be drawn from the correlation coefficient associated with the linear equation is that there's a weak negative correlation between the variables

Learn more about data on:

brainly.com/question/4219149

#SPJ1

7 0

2 years ago

You invest an initial $2,000 in an account that has an annual interest rate of 6%, compounds daily. How much money will you have

Vlad1618 [11]

2000(1+0.06/360)^15*360=??

5 0

3 years ago

If (a^2 + b^2) (x^2 +y^2) = (ax + by )^2 then which relation is true?

NikAS [45]

Answer:

Step-by-step explanation:

Hello,

Let's develop both expressions and try to re factorise.

$\forall (a,b,x,y) \in \mathbb{R}^4 \\ \\ (a^2+b^2)(x^2+y^2)=(ax+by)^2\\ \\(ax)^2+(ay)^2+(bx)^2+(by)^2=(ax)^2+(by)^2+2(ax)(by)\\ \\ (ay)^2+(bx)^2-2(ay)(bx)=0\\ \\ (ay-bx)^2=0\\ \\ ay-bx=0\\ \\ \boxed{\sf \bf ay=bx}$