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2 years ago

12

Ryan can eat 3 hot dogs in 2 minutes at this rate, how many hot dogs could he eat in 7 minutes

2 answers:

Lelechka [254]2 years ago

8 0

Answer:

I do believe it it would be 10 hot dogs

Step-by-step explanation:

I got this by drawing a table

# of hot dogs # of minutes

3 2

6 4

9 6

10 7

tiny-mole [99]2 years ago

3 0

You could layout the question like this:

3 hot dogs = 2 minutes
x hot dogs = 7 minutes

In order to get from 2 minutes to 7 minutes, we could divide by 2 and then multiply by 7. So let’s do that to both sides.

3/2 = 1.5 x 7 = 10.5.

Therefore,

3 hot dogs = 2 minutes
10.5 hot dogs = 7 minutes

You might be interested in

10| x – 4| - 3 > 47

Andrew [12]

Answer:

x < -1 or x > 9

Step-by-step explanation:

$10|x-4|-3>47$

Add 3 to both sides.

$10|x-4|-3+3>47+3$

Simplify.

$10|x-4|>50$

Divide both sides by 10.

$\frac{10|x-4|}{10} >\frac{50}{10}$

Simplify.

$|x-4|>5$

Apply absolute value rule: If $|u|>a,a>0$ then $u < -a$ or $u>a$

$x-4$ or $x-4 >5$

$x-4$

Add 4 to both sides.

$x-4+4$

Simplify.

$x$

$x-4>5$

Add 4 to both sides.

$x-4+4>5+4$

Simplify.

$x>9$

Combine the intervals.

$x$ or $x>9$

8 0

3 years ago

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

6 0

3 years ago

I need help on this problem. Does anyone know how to complete it? Please and thank you :)

ale4655 [162]

I think it's 5 because i remember learning that it's gonna be half of what the bottom is

4 0

3 years ago

Factor 20a + 45.............................................

Delvig [45]

Answer:5(4a+9)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Andreas93 [3]

Answer: (x2 – 1) – 5(x – 1)

7 0

3 years ago

Read 2 more answers