Cabs charge $3.60 Flag Fall plus $2.19 per km. (a) Write down an equation to calculate the cost, C, of cab fare for a distance d
1 answer:
(a) C = 3.60+2.19d
(b) The cab fare will be $36.45 for a distance of d kilometers.
Step-by-step explanation:
Given,
Charges of Flag Fall = $3.60
Per kilometer charges = $2.19
Let,
d be the total distance.
C be the total cost for d kilometers.
(a) Write down an equation to calculate the cost, C, of cab fare for a distance d kilometres.
Total cost = Charges of flag fall + Per kilometer charges * distance
C = 3.60+2.19d
(b) Use your equation to calculate the cost of cab fare for a distance of 15 km.
Here,
d=15
Putting in Equation
The cab fare will be $36.45 for a distance of d kilometers.
Keywords: distance, addition
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Answer: The answers is alternate interior angles.
Step-by-step explanation: First of all, the questions marks given in the figure are renamed in the attached figure as (a), (b), (c) and (d).
For (a): Since AC is parallel to A'C' and A'D is a transversal for these two parallel lines, so, ∠CDB' = ∠B'A'C', because these are alternate interior angles.
For (b): Since BC is parallel to B'C' and A'B' is a transversal, so ∠BEB' = ∠A'B'C', because these are alternate interior angles.
For (c): Since AB is parallel to A'B' and AD is a transversal, so ∠BAC = ∠CDB', because these are alternate interior angles.
For (d): Since AB is parallel to A'B' and BE is a transversal, so ∠ABC = ∠BEB', because these are alternate interior angles.
Thus, all the questions marks are the reasons that the given angles are equal because they are alternate interior angles.
Answer:
You can use either of the following to find "a":
- Pythagorean theorem
- Law of Cosines
Step-by-step explanation:
It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.
I find it reasonably convenient to find the length of x using the sine of the 70° angle:
x = (15 ft)/sin(70°)
x ≈ 15.96 ft
That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.
__
Consider the diagram below. The relation between DE and AE can be written as ...
DE/AE = tan(70°)
AE = DE/tan(70°) = DE·tan(20°)
AE = 15·tan(20°) ≈ 5.459554
Then the length EC is ...
EC = AC - AE
EC = 6.3 - DE·tan(20°) ≈ 0.840446
Now, we can find DC using the Pythagorean theorem:
DC² = DE² + EC²
DC = √(15² +0.840446²) ≈ 15.023527
a ≈ 15.02 ft
_____
You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)
DC² = AD² + AC² - 2·AD·AC·cos(A)
a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635
a = √225.70635 ≈ 15.0235 . . . feet
