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3 years ago

HELP!!!! Order the lengths of the sides of ABC from greatest to least

Mathematics

1 answer:

azamat3 years ago

6 0

The comparison of the lengths of the sides based on the angles are given

by sine rule.

The lengths of the sides from greatest to least are;

<u>BC > AC > AB</u>

Reasons:

The given parameters are;

∠A = 78°, ∠B = 56°, ∠C = 46°

By sine rule, we have;

$\displaystyle \frac{AC}{sin(56^{\circ})} = \frac{AB}{sin(46^{\circ})} = \frac{BC}{sin(78^{\circ})}$

Given that, sin(46°) ≈ 0.719, sin(56°) ≈ 0.829, sin(78°) ≈ 0.978, we have;

sin(78°) > sin(56°) > sin(46°)

Therefore, applying equal proportions of similar ratios, we have;

Learn more about sine rule here:

brainly.com/question/7591047

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3 years ago

The function f(x)=Acosh(Cx)+Bsinh(Cx). I need help determining the value of constant C>0 when the second derivative f''(x)=25

Rus_ich [418]

Answer:

C = 5.

Step-by-step explanation:

First, you need to remember that:

For the function:

h(x) = Sinh(k*x)

We have:

h'(x) = k*Cosh(k*x)

and for the Cosh function:

g(x) = Cosh(k*x)

g'(x) = k*Cosh(k*x).

Now let's go to our problem:

We have f(x) = A*cosh(C*x) + B*Sinh(C*x)

We want to find the value of C such that:

f''(x) = 25*f(x)

So let's derive f(x):

f'(x) = A*C*Sinh(C*x) + B*C*Cosh(C*x)

and again:

f''(x) = A*C*C*Cosh(C*x) + B*C*C*Sinh(C*x)

f''(x) = C^2*(A*cosh(C*x) + B*Sinh(C*x)) = C^2*f(x)

And we wanted to get:

f''(x) = 25*f(x) = C^2*f(x)

then:

25 = C^2

√25 = C

And because we know that C > 0, we take the positive solution of the square root, then:

C = 5

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3 years ago