ANSWERS
1. B
2. B
3. A
4. C
5. B
6. A
WORKINGS
QUESTION 1
7 pt 1c + 4pt 1c (pt for Pints; c for Cups)
= 7pt + 1c + 4pt + 1c
Collecting like terms
= 7pt + 4pt + 1c + 1c
= 11pt + 2c
1pt = 2c
Therefore,
11pt 2c = 11pt + 1pt
= 12 pt (OPTION B)
QUESTION 2
7c 3fl oz – 3c 4fl oz (c for Cups; oz for Ounce)
= 7c + 3fl oz – (3c + 4fl oz)
= 7c + 3fl oz – 3c – 4fl oz
Collecting like terms
= 7c – 3c + 3fl oz – 4fl oz
= 4c – 1fl oz
1c = 8fl oz
Therefore, 4c can be rewritten as 3c + 8fl oz
4c – 1fl oz = 3c +
8fl oz – 1fl oz
= 3c + 7fl oz
= 3c 7fl oz (Option B)
QUESTION 3
4lb 4oz – 2lb 7oz (lb for Pounds; oz for Ounce)
= 4lb + 4oz – (2lb + 7oz)
= 4lb + 4oz – 2lb – 7oz
Collecting like terms
= 4lb – 2lb + 4oz – 7oz
= 2lb – 3oz
1lb = 16oz
Therefore, 2lb can be rewritten as 1lb + 16oz
2lb – 3oz = 1lb + 16oz – 3oz
= 1lb + 13oz
= 1lb 13oz (Option A)
QUESTION 4
7T 200lb – 2T 700lb (T for Tons; lb for Pounds)
= 7T + 200lb – (2T + 700lb)
= 7T + 200lb – 2T – 700lb
Collecting like terms
= 7T – 2T + 200lb – 700lb
= 5T – 500lb
1T = 2,000lb
Therefore, 5T can be rewritten as 4T + 2,000lb
5T – 500lb = 4T + 2,000lb – 500lb]
= 4T + 1,500lb
= 4T 1,500lb (Option C)
QUESTION 5
7 d 12 h – 5 d 18 h (d for Days; h for Hours)
= 7d + 12h – (5d + 18h)
= 7d + 12h – 5d – 18h
Collecting like terms
= 7d – 5d + 12h – 18h
= 2d – 6h
1d = 24h
Therefore, 2d can be rewritten as 1d + 24h
2d – 6h = 1d + 24h – 6h
= 1d + 18h
= 1d 18h (Option B)
QUESTION 6
3 h 42 min – 1 h 12 min (h for Hours; min for Minutes)
= 3h + 42min – (1h + 12min)
= 3h + 42min – 1h – 12min
Collecting like terms
3h – 1h + 42min – 12min
= 2h + 30min
= 2h 30min (Option A)
Answer:
A ) The Equation of the circle (x−10)2+(y−24)2=676
Step-by-step explanation:
step1:-
The equation of the circle whose center is (a,b) and radius r is
in this circle equation centre is (g,f) = (10,24)
and formula of radius of a circle is
r =
Step2:-
The Equation of the circle (x−10)^2+(y−24)^2=676
Answer:
A; Orthocenter
Step-by-step explanation:
A) 0.2076
B) 0.6786
C) 0.3214
Explanation:A)
B) We find the probability that X=6, 7, or 8:
C) P(X≤5) = 1-P(X>5) = 1-0.6786 = 0.3214
The answer is: 13 units.
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Each side of the park is 13 units long.
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(Assuming hexagonal shape will have 6 (SIX) sides of equal length).
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Explanation:
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Let us assume you meant to write that the: "...new park, in the shape of hexagon, will have 6 (six) side of equal length."
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From the coordinates given, we can infer that this is a "regular" hexagon.
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Here is one way to solve the problem: Find the length of ONE side of the hexagon.
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Let us choose the following coordinates: (18,0), and (6.5, 5). Let the distance between these points , which would equal ONE side of our hexagon, represent "c", the hypotenuse of a right triangle. We want to solve for this value, "c".
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Let the distance on the x-axis, from (6.5, 0) to (18.5, 0); represent "b", one side of a right triangle.
→ We can solve for "b" ; → b = 18.5 - 6.5 ; → b = 12 .
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Let the distance from (6.5, 0) to (6.5, 5) ; represent "a"; the remaining side of the right triangle.
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→ a = y₂ - y₁ = 5 - 0 = 5 ;
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{Note: We choose the particular coordinates, including "(6.5, 0)", because the distances between the coordinates chosen form a "right triangle"; (with "c", representing a "hypotenuse", or "slanted line segment"; which would be also be "ONE line segment of the given regular hexagon", which is our answer, because each line segment is the same values, so we only have to find the value of ONE line segment, or side, of the hexagon.).
When considering the given coordinates: "(6.5, 5)", and "(18.5, 0)", a "right triangle" can be formed at the coordinate, "(6.5, 0),
By choosing this particular letters (variables) to represent the sides of a "right triangle", we can solve for the "hypotenuse, "c", using the Pythagorean theorem for the sides of a right triangle:
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→ a² + b² = c² ; in which "c" represents the hypotenuse of the right triangle
and "a" represents the length of one of the other sides; and "b" represents the length of the remaining side. (Note: All triangles have three sides).
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We have: a = 5 ; b = 12 ; → Solve for "c" ;
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→ a² + b² = c² ; ↔ c² = a² + b² ; Plug in the known values for "a" & "b" ;
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→ c² = a² + b² ; → c² = 5² + 12² ;
→ c² = 25 + 144 = 169 ; → c² = 169 ;
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→ Take the square root of each side; to isolate "c" on one side of the equation; and to solve for "c" ;
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→ c² = 169 ; √(c²) = √(169) ; → c = ± 13;
→ ignore the negative value; since the side of a polygon cannot be a negative number;
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→ c = 13 ; The answer is: 13 units.
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