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2 years ago

How to find the derivative of this?

Mathematics

1 answer:

dimulka [17.4K]2 years ago

4 0

I cant see all the question but this is how you find the derivative of your function using the product rule.

Here you use the extension of the product rule to 3 factors which we'll write as:-

f(x), g(x) and h(x):-

Derivative = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)

(3x - 1)(x + 4)(2x - 1)

derivative = 3(x + 4)(2x - 1) + (3x + 1)(1)(2x - 1) + (3x - 1)(x + 4)(2)

= 3(x + 4)(2x - 1) + (3x + 1)(2x - 1) + 2(3x - 1)(x + 4)

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3 years ago

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2 years ago

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ok so if you have 3 and 1/4 pizza and want to subtract 3/4 out of it,and get 3x more slices then you had from a friend after sub

Brilliant_brown [7]

Answer:

You have 2 and 1/2 pizzas and your friend has 5/6 of a pizza.

Step-by-step explanation:

You have 3 1/4 pizzas. If you get rid of 3/4 of a pizza, you will end up with 2 2/4 pizzas.

To make it easier, you can just change the values into improper fractions: 3 1/4 = 13/4

13/4 - 3/4 = 10/4

10/4 = 2 2/4 pizzas

(2 2/4 = 2 1/2)

Anyways! You now have 2 and a half pizzas. You have 3 times the amount of pizza that your friend has. In other words, your friend has 1/3 the amount of pizza that you have. You can solve for this in two ways, whichever way you solve it is up to you.

You can solve this problem by dividing the amount of food you have by 3: $\frac{2\frac{1}{2} }{3}$ which equals 0.8333... (the 3 is repeating). But you can just keep your final answer in fraction form if you want to: $\frac{2.5}{3} = \frac{5}{6}$
You can solve this problem by multiplying 1/3 to the amount of food you have: $\frac{1}{3} * 2 \frac{2}{4}\\=\frac{10}{12}\\ =\frac{5}{6}$

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2 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

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3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

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