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3 years ago

Opal plots point D at (Negative 3 and one-half, one-half). Which graph shows the location of point D?

Mathematics

2 answers:

Doss [256]3 years ago

6 0

Answer:

Step-by-step explanation:

IM DORA

tekilochka [14]3 years ago

6 0

Answer:

ur welcome ......

Step-by-step explanation:

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Round 68.09 to 2 Sf please

marta [7]

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3 years ago

Find the angle measurements and side lengths for the triangle below

rewona [7]

Answer:

Step-by-step explanation:

Sin(A) = 5/7

Sin(A) = 0.7143

A = sin-1(0.7143

A = 45.58

B = 180 - 45.58 - 90

B = 44.43

C = 90

c = 7

a = 5

b^2 = c^2 - a^2

b^2 = 7^2 - 5^2

b^2 = 49 - 25

b^2 = 24

b = 4.899

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3 years ago

For how many values of x between -5 and 5 does f(x) = 1

inysia [295]

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Step-by-step explanation:

3 0

2 years ago

Another name for the x axis

aalyn [17]

Answer:

axis of abscissas

Step-by-step explanation:

8 0

3 years ago

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!

<u><em>Hope that helps!</em></u>

6 0

3 years ago