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2 years ago

Help! Giving BRAINLIEST!!!!!

Mathematics

1 answer:

GrogVix [38]2 years ago

8 0

Answer:

They did not properly distribute the radicals

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7, 12, 10, 18, 13, 23, 29, 15, 16, 18, 15, 12, 20<br> find the outlier. explain your answer.

Naddik [55]

Answer:

Step-by-step explanation:

the rest of the numbers are closer to each other but if you plot these you see that 29 doesnt belong

4 0

2 years ago

Incorrect

GaryK [48]

Answer:

m is the right answer

Step-by-step explanation:

no step hope yu mark me as brainliest

7 0

2 years ago

!!please, please help me!!

grigory [225]

Answer:

Length: 12

Width: 2 $\sqrt{10\\}$

Area: 24 $\sqrt{10\\}$

Step-by-step explanation:

We first find the points of C and D to find the length.

C: (-4,-5)

D: (2,-3)

The distance is 2 $\sqrt{10\\$ .

Now for the height.

A: (2,9)

D: (2,-3)

The distance is 12.

Area: 2 $\sqrt{10\\$ ×12 = 24 $\sqrt{10\\}$

6 0

2 years ago

Simplify 4 (3 -- 4x

Burka [1]

Answer:

12-16x

Step-by-step explanation:

4(3-4x)

12-16x

7 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