Answer:
1 meter equals 100 centimeters, to find out how many centimeters you want to take 0.24*100= 2.4,
Therefore; the computer tablet is 24 centimeters long.
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:
(3, 4) and (5, 12)
Step-by-step explanation:
The points where the two curves intersect are the solutions to the system of equations:
(3, 4) and (5, 12)
Answer:
p = -2/5 or 3/8
Step-by-step explanation:
|5 − 3p| + 9 = 13p + 8
|5 − 3p| = 13p − 1
If 5 − 3p is positive:
5 − 3p = 13p − 1
6 = 16p
p = 3/8
If 5 − 3p is negative:
-(5 − 3p) = 13p − 1
-5 + 3p = 13p − 1
-4 = 10p
p = -2/5
Alternatively, we can square both sides of the equation:
(5 − 3p)² = (13p − 1)²
25 − 30p + 9p² = 169p² − 26p + 1
0 = 160p² + 4p − 24
0 = 40p² + p − 6
0 = (5p + 2) (8p − 3)
p = -2/5 or 3/8
Answer: The greatest number of pages Kenji can decorate = 3
Step-by-step explanation:
Given: Total heart stickers = 15
Total star stickers =12
If all the papers identical, with the same combination of heart and star stickers and no stickers left over.
Then the greatest number of pages Kenji can decorate = GCD(15,12) [GCD=greatest common divisor]
Since 15 = 3 x 5
12=2 x 2 x 3
GCD(15,12) =3
Hence, the greatest number of pages Kenji can decorate = 3
