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3 years ago

Help me out with this question (geometry)

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

3 0

Answer:

it will be (5,3) it think

Step-by-step explanation:

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_____ lateral faces of a rectangular pyramid are congruent. All of the Opposite Adjacent

Juli2301 [7.4K]

Opposite lateral faces of a rectangular pyramid are congruent.

I think is OPPOSITE .

7 0

3 years ago

Read 2 more answers

If p(a|b) = 0.35, p(b) = 0.75 and p(a) = 0.44 are the events a and b independent ?

MatroZZZ [7]

Hello :
note 1 :
p(A|B) = p(A∩B)/ p(B)

note 2 :
 A and B independent: p(A∩B) = p(A)×p(B)
calculate : p(A∩B)
p(A∩B) = p(A|B)×p(B)
p(A∩B) = 0.35 ×0.75 = 0.2625
but : p(A)×p(B) = 0.44×0.75 = 0.33
conclusion :
the events A and B are not independent ?

6 0

3 years ago

Help its due in 15 mins

Anton [14]

Answer:

$\sqrt{44} = 2\sqrt{11}$

Step-by-step explanation:

Apply pythagorean theorem:

ABD is a right triangle with angle ADB = 90

hence AD = $\sqrt{AB^{2} - BD^{2} }$ where BD = 10 because it is half of BC

AD = $\sqrt{12^{2}-10^{2} }$ = $\sqrt{44} = 2\sqrt{11}$

8 0

3 years ago

What is the interest rate, if $48 interest is earned on a depost of $400?

Setler79 [48]

Answer:

im looking for anwser

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

PLEASE HELP MEEEE HURRRY!!! :)

sammy [17]

Answer:

Option D

Step-by-step explanation:

We are given the following equations -

$\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}$

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

$\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}$ First, I swapped the first and third rows.

$\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}$ Leading coefficient of row 2 canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}$ The start value of row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}$ Matrix rows 2 and 3 were swapped.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}$ Leading coefficient in row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}$

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

$\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}$

The positioning of the zeros indicated that there was no solution!

Hope that helps!

6 0

3 years ago