ALGEBRA determine the angle measure in the polygon.a. 30°, 150°, 30°, 150°b. 30°,30°,30°,30°c. 45°,225°,45°,255°D. 15°,75°,15°,7
Veseljchak [2.6K]
We know that consecutive angles in a parallelogram are supplementary, as they are consecutive interior angles between parallel lines.
Then, if we add the measure of x° and the measure of 5x° we get 180°. Then, we can write this expression and solve for x:
Then, the angle x has a measure of 30° and the angle 5x has a measure of 5*30=150°.
Answer: a. 30°, 150°, 30°, 150°
Answer:
neither; perpendicular; parallel; neither; neither
Step-by-step explanation:
parallel line is where the 2 slopes have same gradient and perpendicular means the two lines have a -1 slope when producted
Topic: coordinate geometry
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The correct answer is the top right! :)
Answer:
Step-by-step explanation:
Vertical Asymptote: x=2Horizontal Asymptote: NoneEquation of the Slant/Oblique Asymptote: y=x 3+23 Explanation:Given:y=f(x)=x2−93x−6Step.1:To find the Vertical Asymptote:a. Factor where possibleb. Cancel common factors, if anyc. Set Denominator = 0We will start following the steps:Consider:y=f(x)=x2−93x−6We will factor where possible:y=f(x)=(x+3)(x−3)3x−6If there are any common factors in the numerator and the denominator, we can cancel them.But, we do not have any.Hence, we will move on.Next, we set the denominator to zero.(3x−6)=0Add 6 to both sides.(3x−6+6)=0+6(3x−6+6)=0+6⇒3x=6⇒x=63=2Hence, our Vertical Asymptote is at x=2Refer to the graph below:enter image source hereStep.2:To find the Horizontal Asymptote:Consider:y=f(x)=x2−93x−6Since the highest degree of the numerator is greater than the highest degree of the denominator,Horizontal Asymptote DOES NOT EXISTStep.3:To find the Slant/Oblique Asymptote:Consider:y=f(x)=x2−93x−6Since, the highest degree of the numerator is one more than the highest degree of the denominator, we do have a Slant/Oblique AsymptoteWe will now perform the Polynomial Long Division usingy=f(x)=x2−93x−6enter image source hereHence, the Result of our Long Polynomial Division isx3+23+(−53x−6)
The answer is f(x)‐¹=(+/-)x+25+5
