Six hundred thirty seven thousand eight hundred fifty three
From the z-table and at p67 or 0.67 (the decimal values in the table),
Z =0.44
See the attached photo (at the point where 0.6700 appears)
Complete the square to rewrite the quadratic:
2 <em>x</em>² + 3 <em>x</em> + 5 = 2 (<em>x</em>² + 3/2 <em>x</em>) + 5
... = 2 (<em>x</em>² + 3/2 <em>x</em> + 9/16 - 9/16) + 5
... = 2 (<em>x</em>² + 3/2 <em>x</em> + (3/4)²) + 5 - 9/8
... = 2 (<em>x</em> + 3/4)² + 31/8
Any real number squared becomes non-negative, so the quadratic expression has a minimum value of 31/8, which is greater than 0, and so there are no (real) <em>x</em> for which <em>y</em> = 0.
Divide 6 2/3 by 4 and you get an x value of 1.67 to double check the answer we can multiply 1 2/3 by 4 and we get 6 2/3
This is geometry through and through. Plus a little trig thrown in for fun. If you inscribe an equilateral triangle inside a circle and the triangle has side lengths of 12, you have part of what you need to use Pythagorean's Theorem to find the hypotenuse of the triangle which is also the radius of the circle. First, use the formula 360/3 to find that the central angle measure of each angle INSIDE a triangle is 120. So you have 3 triangles within the large one, each with a top angle of 120 and a base of 12. If you extract that one triangle and then split it in half, you have a right triangle with a base of 6. This is a 30-60-90 triangle and this is important so you can check your work. Now use the apothem formula for a right triangle as it relates to a side in an equilateral triangle, which is a = sqrt3/6 * s. Our values are a = sqrt3/6(12) which simplifies to 2sqrt3. That's our apothem. If you're familiar with a 30-60-0 triangle, you could check this to see it's correct. Now you have the base leg of 6 and the height of 2sqrt3, now use Pythagorean's Theorem to find the hypotenuse, which is also the radius of the circle. This was really a difficult one to explain.
