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Anit [1.1K]
3 years ago
14

Look at the picture plz

Mathematics
2 answers:
Shkiper50 [21]3 years ago
7 0

Answer: y=4x-9

Step-by-step explanation: what you are supposed to do is use the slope intercept formula. y=mx+b

VLD [36.1K]3 years ago
5 0

Answer:

y=4x-9

Step-by-step explanation:

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A pair of equations is shown below:
Elan Coil [88]
So, both equations are essentially linear equations.
Linear equations are written in the format y = mx+b, where m represents the slope/slope intercept and b represents the y-intercept.
---------------------------------------------------------------------------------------------------------------
Part A
the slope and y-intercept

y = mx +b
m - the slope; b- y-intercept

therefore;

y = 6x - 4
m = 6; b = -4

y = 5x - 3
m = 5; b = -3

The coordinates of the point where the lines are crossed are the solution to the system of linear equations.

How to graph the lines:
y = 6x - 4
y-intercept (0; -4)
for x = 1 ⇒ y = 6 · 1 - 4 = 6 - 4 = 2 ⇒ (1; 2)

y = 5x - 3
y-intercept (0; -3)
for x=1 ⇒ y = 5 · 1 - 3 = 5 - 3 = 2 ⇒ (1; 2)

***look at the img for graph reference***

Part B: x = 1; y = 2

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Brainliest answer ------- Which transactions should be recorded in the Withdrawal column of a check register? Choose all answers
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Why shouldn’t all ratios be simplified
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What is 10 more than 88
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Write the given expression in terms of x and y only.<br> sin(sin−1(x) + cos−1(y))
yan [13]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2308127

_______________


Write the expression below in terms of x and y only:

(I'm going to call it "E")

\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}


Let

\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}


so the expression becomes

\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}


•   Finding \mathsf{sin\,\alpha:}

\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}


•   Finding \mathsf{cos\,\alpha:}

\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}


because \mathsf{cos\,\alpha} is positive for \mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}


•   Finding \mathsf{cos\,\beta:}

\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}


•   Finding \mathsf{sin\,\beta:}

\mathsf{cos^2\,\alpha=y^2}\\\\&#10; \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ &#10;\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}


because \mathsf{sin\,\beta} is positive for \mathsf{\beta\in [0,\,\pi].}


Finally, you get

\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry</em>

6 0
3 years ago
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