-5.87 then -33 then 0.6 then 17
There are several conditions where triangles can be proved similar:
AA - where two of the angles are same.
SAS - where two sides of a triangle compare to the corresponding sides in the other are in same proportion, and the angle in the middle are equal.
SSS - Where all sides in a triangle and the corresponding sides are in the same proportion.
In the case above, we can only use the method of SAS, as only two sides of the triangles are given.
<HMG = <JMK (vertically opposite angles)
HM/MK = 8/12 = 2/3
GM/MJ = 6/9 = 2/3
As the two sides of a triangle comparing to the corresponding sides in the other are in same proportion, and the angle in the middle are equal, the above triangles are similar, with the prove of SAS.
Therefore, the answer is C.yes by SAS.
Hope it helps!
Answer: I think it is greater than
Step-by-step explanation:
Because positive - negative = positive
Answer:
Kara should plot points where the arcs intersect above and below the line segment.
Step-by-step explanation:
The bisection of a line segment is the dividing of the line segment into two equal parts. To bisect a line segment, you have place your compass on the endpoints and measure a distance greater than half of the segment. The point of intersection of the arcs both above and below the segment is then joined thereby bisecting the line.
Since Kara has already drawn the two arcs which are greater that half of the length, all Kara needs to do is plot points where the arcs intersect above and below the line segment.
Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as
<span>(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). </span>
<span>Next, complete the square. </span>
<span>x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² </span>
<span>Also, for the x in the numerator </span>
<span>x = x + 1/2 - 1/2. </span>
<span>So </span>
<span>(x² + 2x + 3)/(x² + x + 3) = 1 + (x + 1/2)/[(x + 1/2)² + (√(11)/2)²] - 1/2/[(x + 1/2)² + (√(11)/2)²]. </span>
<span>Integrate term by term to get </span>
<span>∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C </span>
<span>b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. </span>
<span>∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C </span>
<span>= 2 √(x) ln(x) - √(x) + C. </span>
<span>c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. </span>
<span>∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C </span>
<span>= arctan(e^x) + C.</span>
