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2 years ago

What are the slope and the y-intercept of the line shown in the graph?

Mathematics

2 answers:

mr Goodwill [35]2 years ago

7 0

Answer:

the answer is A y_ intercept =2 and slope =-3

pickupchik [31]2 years ago

7 0

A. Y =2 slope =-3 hope this helps

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3 years ago

Determine the measure of ∠Y. images attached

Gwar [14]

Answer:

<h3>The answer is option C</h3>

Step-by-step explanation:

Since it's a right angled triangle we can use trigonometric ratios to find the value of ∠Y

To find the value of ∠Y we use tan

We have

$\tan(∠Y) = \frac{25}{20} \\ \tan(∠Y) = \frac{5}{4} \\ ∠Y = { \tan^{ - 1} } \frac{5}{4} \\ ∠Y = 51.34019...$

We have the final answer as

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3 years ago

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Which set of angles could be the interior angles of a triangles

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Step-by-step explanation:

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2 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago