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2 years ago

PLESSE HRLP IM BEGGING U ILL MARK U

Mathematics

1 answer:

Andrew [12]2 years ago

7 0

Answer:
y = 1/2x + 14

Explanation:
y = 1/2x + 14 is perpendicular to y = -x = 18 and passes through the slope (-6, 11)

Tip:
Desmos (the app) helps a lot with graphing problems.

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Help help help help pls

Monica [59]

Answer:

$\theta = 82.2^\circ$

Step-by-step explanation:

$Let \ \angle BAC = \theta$

Opposite = BC ,

Adjacent = AB = x = 3 ,

Hypotenuse = AC = y = 22

Using trigonometric ratios.

$sin \theta = \frac{opposite}{hypotenuse }\\\\cos \theta = \frac{adjacent}{hypotenuse}\\\\tan \theta = \frac{opposite}{adjacent}$

Since we have adjacent and hypotenuse we use cosine's ratio

to find the angle.

$cos \theta = \frac{x}{y} = \frac{3}{22} \\\\\theta = cos^{-1} \frac{3}{22} = 82.16 = 82.2^\circ$

6 0

3 years ago

Solve 7 8 and 9 please

VLD [36.1K]

7) y = 4x
8) y = 7x/8
9) y = 25x/3

I think this are the answers.

7 0

3 years ago

Read 2 more answers

Plz help urgent !!!! will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...