14 + 63 = 2*7 + 9*7 = 7*(2+9)
the length of the segment AD is 5 units
Answer:
Midpoint (-2,4)
distance nearest tenth = 8.9
The approximate distance = 9
Step-by-step explanation:
Formulas
PQ midpoint = (x2 + x1)/2, (y2 + y1)/2
distance d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )
Givens
x2 = -4
x1 = 0
y2 = 1
y1 = 7
Solution
M(PQ) = (-4+0)/2, (1 + 7)/2
M(PQ) = -2, 4
The midpoint is -2,4
The distance = sqrt( (4 - 0)^2 + (1 + 7)^2 )
The distance = sqrt(16 + 64)
The distance = sqrt(80)
The distance = 4√5 exactly
The distance = 8.94
The distance = 8.9 To the nearest tenth
Question 2
The distance is rounded to the nearest whole number which is 9.
<h3>
Answer: 138,240 cubic cm</h3>
Explanation:
If x is the side length of a cube, then x^3 is the volume. Think of it like length*width*height, but length = width = height.
It's no coincidence that the cubing operation or cubed exponent ties directly to the concept of things like volumes in cubic cm.
In this case, x = 12 cm is the side length of each small cube. The volume of each small cube is x^3 = 12^3 = 12*12*12 = 1728 cubic cm.
Eighty such small cubes get us a total volume of 80*1728 = 138,240 cubic cm.
Answer:
i). x³ + 9x² + yz - 15
ii). -21m³np - 8p⁵q + mnp + 4mn + 100
Step-by-step explanation:
Question (38)
i). Two expressions are -5x² - 4yz + 15 and x³+ 4x²- 3yz
By subtracting expression (1) from expression (2) we can the expression by addition which we can get expression (1).
(x³+ 4x²- 3yz) - (-5x² - 4yz + 15) = x³ + 4x² - 3yz + 5x² + 4yz - 15
= x³ + 9x² + yz - 15
ii). -15m³np + 2p⁵q - 6m³pn + mnp + 4mn - 10qp⁵+ 100
= (-15m³np - 6m³np) + (2p⁵q - 10qp⁵) + mnp + 4mn + 100
= -21m³np - 8p⁵q + mnp + 4mn + 100
