find the perimeter of ABC with verticles A(-1,2) B(3,-2) and C(-1,-2). round answer to nearest hundredth

Mathematics

1 answer:

nata0808 [166]2 years ago

5 0

Answer:

A po

Step-by-step explanation:

sana makatulong haaha

You might be interested in

Which method do you think is easiest? Factoring, Completing the Square, or the Quadratic Formula?

stich3 [128]

All are easy. There is no way to say thatnone way is easier than the other. Well the answer lies in every individual, it depends on which method they are comfortable with. Anyway there are more method to solve quadratic than the listed ones.

5 0

2 years ago

Jordan is going to share sweet candy with 5 friends. He is going to keep 59 caramel candies. Jordan is going to give 25 candy ca

Solnce55 [7]

He is keeping 34 candy

4 0

2 years ago

Suppose that the trace of a 2×2 matrix a is tr(a)=15 and the determinant is det(a)=50. find the eigenvalues of

IrinaK [193]

Recall that the characteristic polynomial of a 2x2 matrix $\mathbf A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is

$\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}a-\lambda&b\\c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)$

but $\det(\mathbf A)=ad-bc$ and $\mathrm{tr}(\mathbf A)=a+d$ , so the characteristic polynomial for $\mathbf A$ is

$\lambda^2-\mathrm{tr}(\mathbf A)\lambda+\det(\mathbf A)$

We're given that the trace is 15 and determinant is 50, so the characteristic polynomial for the matrix in question is

$\lambda^2-15\lambda+50$

and the eigenvalues are those $\lambda$ for which the characteristic polynomial evaluates to 0.

$\lambda^2-15\lambda+50=(\lambda-5)(\lambda-10)=0\implies\lambda=5,\lambda=10$

5 0

2 years ago

What are the constants in the following expression?<br><br><br><br> +7p +8d +12

nevsk [136]

I’m very confused what are you trying to do here ?

5 0

2 years ago

EASY TRUE OR FALSE!!! WILL GIVE BRAINLIEST!!!!

mixer [17]

<h2>the answer is true. ☺</h2>

7 0

2 years ago

Read 2 more answers