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1 year ago

Answer Part A and Part B for the triangle shown

Mathematics

1 answer:

Tom [10]1 year ago

5 0

Answer:

for Part A

Answer is C.

Step-by-step explanation:

$\tan(a) = \frac{opposite}{adjacent } \\$

Here the angle, a is 60°, Opposite is 20 and adjacent is x.

Therefore we have,

$\tan(60 ) = \frac{20}{x}$

Making x the subject gives us

$x = \frac{20}{ \tan(60) }$

Part B

Value of x to the nearest tenth

$x = 11.547 \\ to \: the \: nearest \: tenth \\ x = 12$

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A bank quotes an interest rate as 0.06341 annual effective yield. What interest rate, compounded monthly, will provide that

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Answer:

0.06164

Step-by-step explanation:

The effective annual rate obtained by compounding nominal annual rate r monthly is ...

eff rate = (1 +r/12)^12 -1

Then the value of r is ...

r = 12×((eff rate) +1)^(1/12) -1)

For the given effective rate, that is ...

r = 12×(1.06341^(1/12) -1) ≈ 0.06164 . . . . nominal annual interest rate

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2 years ago

Find the 2nd Derivative:<br> f(x) = 3x⁴ + 2x² - 8x + 4

ad-work [718]

Answer:

$f''(x)=36x^2+4$

Step-by-step explanation:

Let's start by finding the first derivative of $f(x)= 3x^4+2x^2-8x+4$ . We can do so by using the power rule for derivatives.

The power rule states that:

$\frac{d}{dx} (x^n) = n \times x^n^-^1$

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

$\frac{d}{dx} (3x^4+2x^2-8x+4)$

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)$

Simplify the equation.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)$
$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)$

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)$
$f'(x)=12x^3+4x-8$

Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).

$\frac{d}{d} (f'x) = \frac{d}{dx} (12x^3+4x-8)$

$\frac{d}{dx} (12x^3+4x-8) = ((3)12x^3^-^1 + (1)4x^1^-^1 - (0)8)$

Simplify the equation.

$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4x^0 - 0)$
$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4(1) - 0)$
$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4 )$

Therefore, this is the 2nd derivative of the function f(x).

We can say that: $f''(x)=36x^2+4$

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