Answer:
be able to make it to the meeting tonight but I can tomorrow if you have time can you come to my house and I will be there
◆ Define the variables:
Let the calorie content of Candy A = a
and the calorie content of Candy B = b
◆ Form the equations:
One bar of candy A and two bars of candy B have 774 calories. Thus:
a + 2b = 774
Two bars of candy A and one bar of candy B contains 786 calories
2a + b = 786
◆ Solve the equations:
From first equation,
a + 2b = 774
=> a = 774 - 2b
Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie
◆ Find caloric content:
Caloric content of candy B = 254 calorie
Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
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.
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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
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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
Answer:
Please refer to the attachment