Answer:
a) 5units
b) (4.5, 4)
Step-by-step explanation:
Let the coordinate of P and Q be P(3, 2) and Q(6, 6)
PQ can be gotten using the distance formula;
PQ =√(6-2)²+(6-3)²
PQ = √4²+3²
PQ =√16+9
PQ =√25
PQ = 5units
Midpoint M is expressed as;
M(X, Y) = {(x1+x2)/2, y1+y2/2}
X = x1+x2/2
X = 3+6/2
X= 9/2
X = 4.5
Similarly;
Y = y1+y2/2
Y = 2+6/2
Y = 8/2
Y = 4
Hence the midpoint of PQ is (4.5, 4)
Answer:
2(x + 4) / 6(x² - 3x - 28)
Step-by-step explanation:
Area of a rectangle = length × width
Length = 2/(x² - 3x - 28)
Width = x² - 16/6x - 24
= (x + 4)(x - 4) / 6(x - 4)
= (x + 4) / 6
Area of a rectangle = length × width
= 2/(x² - 3x - 28) × (x + 4) / 6
= 2(x + 4) / (x² - 3x - 28)6
= 2(x + 4) / 6x² - 18x - 168
= 2(x + 4) / 6(x² - 3x - 28)
Area of a rectangle =
2(x + 4) / 6(x² - 3x - 28)
Answer:
b
Step-by-step explanation:
The associative property of addition is shown in equations where the order in which the evaluation is made does not change the result, providing the numbers remain in the same order. Option B demonstrates this.
Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
