All categories

3 years ago

A, B & C form a triangle where Z BAC = 90°.

Mathematics

1 answer:

vazorg [7]3 years ago

7 0

Answer:

Step-by-step explanation:

bc is hypotunuse

therefore 1.2^2+3.9^2=BC^2

4.1(1dp)

You might be interested in

Help me pls I need this by today

asambeis [7]

Answer: r=0.4 its a guess but i dont think thats a negitive slope it looks positive so :P

Step-by-step explanation:

4 0

3 years ago

Triangle ABC has vertices at A(2, 2), B(4,7), and C(6,2). Classify the triangle according to the side lengths.

zmey [24]

Answer:

Isosceles triangle

Step-by-step explanation:

Length AB

√(4 - 2)² + (7 - 2)²

√2² + 5²

√4 + 25

√29

Length AC

6 - 2 = 4

Length BC

√(6 - 4)² + (2 - 7)²

√2² + (-5)²

√4 + 25

√29

Length AB and BC is the same.

Length AC is the longest side.

Therefore, it is an isosceles triangle.

4 0

4 years ago

What is 2 to the second power

kodGreya [7K]

Answer:

Step-by-step explanation:

2 to the second power = 2^2.

2^2 means 2*2 = 4.

The 2 in ^2 is called the 'exponent' .

Note 3^3 = 3*3*3

and 5^5 = 5*5*5*5*5.

6 0

3 years ago

Read 2 more answers

I WILL GIVE BRAINLY PLEASE HELP ITS THE TOP ONE

Anastasy [175]

Answer:

I think it is C

Step-by-step explanation:

Forgive me if I am wrong the reason I think so is I did one like this last week. Sorry if it is wrong trying to remember.

5 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago