X + 2y = 11
2x + y = 31
x = 11 - 2y
2(11 - 2y) + y = 31
22 - 4y + y = 31
3y = - 9
y = - 3
x = 11 - 2(-3)
x = 11 + 6
x = 17
<span>rhoumbus & trapazoid are the only ones.</span>
Answer:
A.) 2(1/2×20×50)+2(1/2×20×10)
I think it's 100. You add all of the angles to =720. then solve for x. What you get for x, you plug into the equation for angle A and you should get 160-60=100.
Answer:
- 6. See solution
- 7. k = 2, k = -6
Step-by-step explanation:
6.
<u>Given equation:</u>
- The sum of the roots is q1 and the product of the roots is q2
Need to show that q1+q2 = 0
<h3>Solution</h3>
<u>Bringing the equation into standard form of ax² + bx + c = 0:</u>
- 2(x + 2)² + p(x + 1) = 0
- 2x² + 8x + 8 + px + p = 0
- 2x² + (p + 8)x + (p + 8) = 0
<u>Sum of the roots: </u>
<u>Product of the roots:</u>
<u>We see that q1 and q2 are opposite numbers, therefore their sum equals zero:</u>
- q1 + q2 = -(p + 8)/2 + (p + 8)/2 = 0
=============================================
7.
<u>Given quadratic equation:</u>
Need to find the possible values of k
<h3>Solution</h3>
<u>When the quadratic equation has equal roots, then its discriminant is equal to zero:</u>
- D = 0
- √b² - 4ac = 0
- √(-k -2)² - 4*1*4 = 0
- √k² + 4k + 4 - 16 = 0
- √k² + 4k - 12 = 0
- k² + 4k - 12 = 0
- k = {-4 ± √4² -4*1*(-12)}/2
- k = (-4 ± √16 + 48)/2
- k = (-4±√64)/2
- k = -2 ± 4
- k = 2, k = -6
