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2 years ago

A teacher spends $150 on school supplies for the classroom, folders are $0,25 each and notebooks are $2.50 each

Mathematics

1 answer:

exis [7]2 years ago

6 0

Answer:

a. x=600

b. y=60

c. x = 150 and y = 45

Step-by-step explanation:

0.25x+2.50y=150

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The maximum amount of weight the truck can carry is 1600lbs. You do not want to exceed 1600lbs, but you can reach 1600lbs. This makes the inequality symbol ≤. This symbol says that the weight load has to be less than 1600lbs, but It can equal 1600lbs. You know that the piano is 400lbs and each box is 50lbs. Since you want to figure out how many boxes you can put on the truck alongside the piano, you can use the inequality: 50x + 400 ≤ 1600.

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3 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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