Answer: B) 115 pounds
Step-by-step explanation:
Because x is the number of days passed, after 21 days, x = 21.
Hope it helps <3
A person has to walk 100 m to go from position X in the north of east direction to the position B and then to the west of Y to r
each finally at position Z. The position Z is situated at the north of X and at a distance of 60 m from X. Find the distance between X and Y.
1 answer:
Answer:
The distance between X and Y is 68m
Step-by-step explanation:
Consider the diagram attached:
A person starts from X, goes to Y, and then finally goes to Z, covering a total of 100m, where Z is located 60m in north of X.
We have to find distance XY.
Let the distance XY be x
Total distance covered by the person = XY + YZ
We know that total distance covered is 100m, so
100m = x + YZ
YZ = (100 - x)m
See the diagram, XZY forms a right angled triangle such that:
Hypotenuse = XY = x
Base = XZ = 60m
Perpendicular = YZ = (100-x)m
Using Pythagoras theorem
Hypotenuse² = Base² + Perpendicular²
x² = (60)² + (100-x)²
Solve the equation for x, we get:
x = 68m
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Answer:
In case (a) car makes 105° turn.
In case (b) car makes 75° turn.
In case (c) car makes 105° turn.
Step-by-step explanation:
Figure is redrawn To explain properly (in attachment)
Given : streets are parallel means ║
,
AB - 4th street , CD - 3rd street and XY - King Ave.
∠XLA = 75°
To find : (a) ∠XLB
(b) ∠LMD (left onto 3rd streat means left of car)
(c) ∠YMD (right means right side of car)
∠XLB + ∠XLA = 180° (Linear Pair = 2 adjacent angles are
supplementary)
∠XLB + 75° = 180°
∠XLB = 180 - 75
∠XLB = 105°
∴ In case (a) car makes 105° turn.
∠LMD = ∠XLA = 75° (Corresponding angles of parallel lines are equal)
∠LMD = 75°
∴ In case (b) car makes 75° turn.
∠YMD + ∠LMD = 180° (Linear Pair = 2 adjacent angles are
supplementary)
∠YMD + 75° = 180°
∠YMD = 180 - 75
∠YMD = 105°
∴In case (c) car makes 105° turn.
Answer:
The side length of the original square was 6 inches
Step-by-step explanation:
we know that
The area of a square is
where
b is the length side of the square
Let
x ---> the length of the original square
The area of the original square is
The length of the smaller square is
The area of the smaller square is
The area of the smaller square is 1/4 the area of the original square
so
solve for x
Multiply by 4 both sides
Solve the quadratic equation by graphing
using a graphing tool
x=2, x=6
see the attached figure
The solution is x=6 in
Remember that the solution must be greater than 3 inches (because Stacey cuts 3 inches off of the length of the square and 3 inches off of the width)
therefore
The side length of the original square was 6 inches
