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3 years ago

Using the Law of Sine or Cosine, solve for the unknown variable (round to the

Mathematics

1 answer:

Daniel [21]3 years ago

6 0

Law of Sines always came easier to me since it was just simple cross multiplication.

The formula consists of a/sinA=b/sinB=c/sinC.

However, this one cannot be used for Law of Sines because one more angle isn't given. So off we go to Law of Cosines.

This one is simpler. The formula is c^2 = a^2 + b^2 − 2ab cos(C)

This one is easier to do. Let's designate all sides a variable. Let's just say 47 is a, 56 is b, and c is x.

Let's fill in the numbers

c^2=47^2+56^2-2(47)(56)cos124

c^2=2209+3136-(5264cos124)

c^2=5245-(-2895.2)

c^2=5245+2895.2

c=90.22

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How much simple interest does $4,678.00 earn in 529 days at a rate of 7.3%? Round to the nearest cent.

statuscvo [17]

Answer:

$5,172.93

Step-by-step explanation:

We're gonna use the simple interest formula: P = A(1 + r(t))

P = final amount

A = starting amount (4,678)

r = rate (0.073)

t = years (529/365)

P = 4,678(1 + 0.073(529/365))

P = 5,172.9324

7 0

2 years ago

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago

Help me please, what are the like terms?

andriy [413]

Step-by-step explanation:

Like terms are are terms whose variables (and their exponents such as the 2 in x 2) are the same. In other words, terms that are "like" each other. Note: the coefficients (the numbers you multiply by, such as "5" in 5x) can be different.

So like in your picture x and x would be like terms... your picture helps as well to identify your like terms because your like terms have the same background color to there particular rectangle.

4 0

3 years ago

What is the answer from "x=2" and graph each line and name two point on the line

Elena L [17]

i dont know srry

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4 0

3 years ago

40 POINTS answer correctly

Alexxx [7]

Since the denominator is the number on the bottom, we need to find a common denominator for a and 5a^2. Since a*5a is 5a^2, we have a common denominator - 5a^2! For the fractions, that means that since a*5a=5a^2, we have to multiply -7 by 5a to get -35a/5a^2-9/5a^2=(-35a-9)/5a^2

7 0

3 years ago

Read 2 more answers