Answer:
64 <em> </em><em>te</em><em> </em><em>quiero</em><em> </em><em>mucho</em><em> </em><em>ygrfrf7rttffrtx</em>
Answer:
Step-by-step explanation:
-7k = 0
Divide both sides by (-7)
k = 0
17) c - 4 = -23
Add 4 to both sides
c - 4 + 4 = -23 + 4
c = - 19
This pattern of question is always coming up. Since we can't easily guess, then let us set up simultaneous equation for the statements.
let the two numbers be x and y.
Multiply to 44. x*y = 44 ..........(a)
Add up to 12. x + y = 12 .........(b)
From (b)
y = 12 - x .......(c)
Substitute (c) into (a)
x*y = 44
x*(12 - x) = 44
12x - x² = 44
-x² + 12x = 44
-x² + 12x - 44 = 0.
Multiply both sides by -1
-1(-x² + 12x - 44) = -1*0
x² - 12x + 44 = 0.
This does not look factorizable, so let us just use quadratic formula
comparing to ax² + bx + c = 0, x² - 12x + 44 = 0, a = 1, b = -12, c = 44
x = (-b + √(b² - 4ac)) /2a or (-b - √(b² - 4ac)) /2a
x = (-(-12) + √((-12)² - 4*1*44) )/ (2*1)
x = (12 + √(144 - 176) )/ 2
x = (12 + √-32 )/ 2
√-32 = √(-1 *32) = √-1 * √32 = i * √(16 *2) = i*√16 *√2 = i*4*√2 = 4i√2
Where i is a complex number. Note the equation has two values. We shall include the second, that has negative sign before the square root.
x = (12 + √-32 )/ 2 or (12 - √-32 )/ 2
x = (12 + 4i√2 )/ 2 (12 - 4i√2 )/ 2
x = 12/2 + (4i√2)/2 12/2 - (4i√2)/2
x = 6 + 2i√2 or 6 - 2i√2
Recall equation (c):
y = 12 - x, When x = 6 + 2i√2, y = 12 - (6 + 2i√2) = 12 - 6 - 2i√2 = 6 - 2i√2
When x = 6 - 2i√2, y = 12 - (6 - 2i√2) = 12 - 6 + 2i√2 = 6 + 2i√2
x = 6 + 2i√2, y = 6 - 2i√2
x = 6 - 2i√2, y = 6 + 2i√2
Therefore the two numbers that multiply to 44 and add up to 12 are:
6 + 2i√2 and 6 - 2i√2
For a triangle to be right angled the largest side squared must equal the sum of the squares on the other 2 sides ( Pythagoras theorem)
1. 101^2 = 99^2 + 20^2 so its is right angled
2. 28^2 is not equal to 26^2 + 12^2 - not right angled
3. 17^2 not euql to 14^2 + 6^2 - not
