All categories

3 years ago

Wilson can eat 50 French fries in 4 minutes. How many French fries can he eat in 28 minutes

Mathematics

1 answer:

lions [1.4K]3 years ago

6 0

Wilson can eat 350 French fries in 28 minutes.

You might be interested in

What is 3/5×6 please answer quickly

ludmilkaskok [199]

The answer is 18/5 and simplify that and u get 3 3/5.

7 0

2 years ago

Read 2 more answers

Someone help pleaseeee

qaws [65]

Answer:

Part A = D

Part B = 5 and 50

Step-by-step explanation:

Part A:

area = 5*4x = 20x

so, $100 \leq 20x \leq 1000$

Part B:

to leave it only by x, you divide both sides by 20

100/20 = 5, 1000/20 = 50

which $5\leq x\leq 50$

7 0

3 years ago

natulia [17]

Answer:

1648

Step-by-step explanation:

5/9 are girls. The 9 represents the total.

Set up a proportion and you have

5/9 = ?/3708

3708/9=412

412x5= 2060 girls

3708-2060= 1648

8 0

3 years ago

Read 2 more answers

Perform the operation (9x^2+6x-7)+(-9x-5)

asambeis [7]

Answer:

9 $x^{2}$ -3x-12

9 $x^{2}$ +6x-7-9x-5

(Group liked terms)

9 $x^{2}$ +6x-9x-7-5

( subtract the numbers ) -7-5= -12

9 $x^{2}$ +6x-9x-12

( add similar elements ) 6x-9x=-3x

9 $x^{2}$ -3x-12

Step-by-step explanation:

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D

Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago