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3 years ago

Can somebody please help me with this

Mathematics

1 answer:

Paul [167]3 years ago

4 0

∠PYG = 137°

Step-by-step explanation:

We know that a straight line is equal to 180°, so by looking at the image we can tell that if we added both of these angles together they would be equal to the straight line.

We can make an equation by adding these two angles together and setting them equal to 180.

$2p+153+6p+91=180$

We can combine like terms.

$8p+244=180$

Subtract 244 from both sides.

$8p=-64$

Divide.

p = -8

Now that we have found what p is equal to, we can plug it in to find the measure of ∠PYG.

$2*-8+153$

= 137

If you wanted to double check this answer, we can plug in -8 for p on the other side and make sure both would be equal to 180.

Like this...

$6*-8+91$

= 43

And we know that...

137 + 43 = 180

So...

∠PYG = 137°

∠HYG = 43°

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

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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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