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makvit [3.9K]
3 years ago
14

Can the measurement from △DEF be determined using only the Law of Cosines? State whether each measurement can be determined usin

g only the Law of Cosines by answering “yes” or “no”.
m∠D =


m∠E =


DE =

Mathematics
1 answer:
quester [9]3 years ago
7 0

9514 1404 393

Answer:

  • ∠D - no
  • ∠E - no
  • DE - yes

Step-by-step explanation:

With a single application of the Law of Cosines, you can only find one of an unknown side or an unknown angle. The other three elements in the 4-variable equation must be specified.

However, a single application of the LoC can be used to find DE. Then, knowing the three sides, either of the unknown angles can be found from an additional application of the LoC.

So, the answer is "it depends." It is yes to all if finding DE first is allowed. It is "no" to the angles if they must be found without finding DE first.

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dilation

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Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their buildup in the atmosphere started destroying the o
melamori03 [73]

Answer:

a) C(2000)=1915

C(2014)=1915

b) C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c) C(t)=-\frac{275}{14}t+\frac{288405}{7}

d) t=2021

e) too early

Step-by-step explanation:

a)

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915    concentration in year 2000

C(2014)=1915      concentration in year 2014

b)

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}

so:

C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64

therefore:

C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c)

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

y-y_{1}=m(x-x_{1})

in this case:

C(t)-C(t_1)=C'(t)(t-t_{1})

so we get:

C(t)-1915=-\frac{275}{14}(t-2000)

and we can now solve for C(t) so we get:

C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915

for a final answer of:

C(t)=-\frac{275}{14}t+\frac{288405}{7}

d)

So next we solve the equation for C(t)=1500 so we get:

1500=-\frac{275}{14}t+\frac{288405}{7}

1500-\frac{288405}{7}=-\frac{275}{14}t

-\frac{277905}{7}=-\frac{275}{14}t

t=2021 \frac{7}{55}

so we will reach that concentration level at the year 2021 approximately.

e)

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.

8 0
3 years ago
How to solve a quadratic formula
Deffense [45]
Well the quadratic formula is just:
x = -b + and - the square root of b^2 - 4ac all divided by 2a. You just plug in numbers and get an answer for x.
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