A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the
KM² = KE² + ME²KM² = 8² + (3√5)² = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM) = 1/2 * 3√5 * (40 + 24) = 1/2 * 3√5 * 64 = 3 x 2.23 * 32 = 214.66 units²
Answer:
19 and 16
Step-by-step explanation:
Let x and y be the two numbers
x+y = 35
x-y =3
Add the two equations together
x+y = 35
x-y =3
---------------
2x = 38
Divide by 2
2x/2 = 38/2
x = 19
Solve for y
x-y = 3
19-y = 3
Subtract 19 from each side
19-19 -y = 3-19
-y = -16
y =16
The rational root (or zero) theorem says a polynomial's rational zeros must have factors of the constant in the numerator and those of the leading coefficient in the denominator.
Here the leading coefficient is 1 and the constant is 5, so our only possibilities for rational roots are 1, -1, 5, -5
We try them each in term
1-3+1+5=6 nope
-1 -3 -1 +5 = 0 yes, x = -1 is a rational zero
5 and -5 don't work either so we're left with
Final answer: x = -1
Answer:
77%
Step-by-step explanation:
Honestly I would just count it
Answer:
The Second One I Think
Step-by-step explanation:
By what I understand that would be the one that is in standard form
