All categories

2 years ago

Which of the following are factors of 60 that have a sum of 17?

Mathematics

1 answer:

Mashutka [201]2 years ago

6 0

This question doesn’t have any options but factors of 60 that have a sum of 17 would be 5 and 12.

You might be interested in

Hope someone can help me

Advocard [28]

Hi!

Let's find out how much he has to pay for each of the deals.

Deal 1

£1.75 per mile.

192 x 1.75 = 336

£336

Deal 2

15% off of £680

$\frac{x}{680} \frac{15}{100}$

680 x 15 = 10200

10200/100 = 102

680 - 102 = 578

£578

Deal 3

£90 per day - 1/10

90 x 7 = 630

1/10 of 630 = 63

630 - 63 = 567

£567

The cheapest deal would be Deal 1

Hope this helps! :)

-Peredhel

5 0

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).