All categories

3 years ago

Mhanifa please help with this I find it super difficult! Thanks! I will mark brainliest! Random answers will be reported !

Mathematics

1 answer:

Artyom0805 [142]3 years ago

4 0

Answer:

The point that is equidistant to all sides of a triangle is called the incenter.

The incenter is located at the intersection of bisectors of the interior angles of a triangle.

The point that is equidistant to all vertices of a triangle is called the circumcenter.

The circumcenter is located at the intersection of perpendicular bisectors of the sides of a triangle.

See the attachment

The blue lines and their intersection shows the incenter.

The red lines and their intersection shows the circumcenter.

As we see the red point- the circumcenter is closer to vertex B.

You might be interested in

How do you use a model to show 3.1 = 3.10

n200080 [17]

Because every number with the decimal point have a zero infinit so it's not matter they're are same

5 0

3 years ago

HELP ME WHICH ANSWER

SIZIF [17.4K]

Answer:

Step-by-step explanation:

30 : 13 = 30/13

8 0

2 years ago

Read 2 more answers

Plz tell me this solution it's very urgent

a_sh-v [17]

Step-by-step explanation:

here is fig 15

x=36 (v.o.a are equal)

36+y=180

y=180-36=144

y=z=144 vertically

3 0

3 years ago

Please help me i rlly need help lol i have to written

MissTica

Take the picture clearly because I can’t read nothing

6 0

3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago