How would I write #3

Mathematics

2 answers:

Rainbow [258]3 years ago

7 0

Yes, if you know that every 1 quart is equal to 4 cups then the ratio is 1:4. To find out how many cups are in 9 quarts, multiply both sides of the ration by 9. The answer would be 9:36, so there are 36 cups in 9 quarts.

Shkiper50 [21]3 years ago

4 0

It would be (9,36). If you look at the graph, the dots just go one right one up, so if u just keep placing more dots in a line till you get to 9, it'll land on 36. Also another way to do it is you could look at the table (the one with numbers placed in the boxes), you can see that the pattern goes quarts times 4 makes the number of cups. For example, 1 times 4. So if you did9 times 4 that would also be 36.

You might be interested in

Please help with this problem

HACTEHA [7]

Answer:

x = 2

Step-by-step explanation:

Multiply both sides by (x-5) and expand:

(x+4)/(x-5)*(x-5) = -2(x-5)

x+4 = -2x+10

Subtract 4 from both sides:

x+4-4 = -2x+10-4

x = -2x+6

Add 2x to both sides:

x+2x = -2x+6+2x

3x = 6

Divide both sides by 3:

3x/3 = 6/3

x = 2

7 0

3 years ago

Read 2 more answers

Need help please :( What is the second term of (s+v)^5?

Citrus2011 [14]

First compute the coefficient like this:
$C_5^1=\frac{5!}{1!(5-1)!}=\frac{5!}{4!}=\frac{5\times4!}{4!}$
Simplifying the fraction over 4! we get:
$C_5^1=5$
and the variables are $s^4v$ . So answer $5s^4v$ .

The correct answer is C then.

3 0

3 years ago

An automobile firm wants to purchase either machine A or machine B. The intial cost of machines A and B are $40,000 and $50,000,

Fynjy0 [20]

Answer:

C. $118,000 and $110,000, respectively.

Step-by-step explanation:

Please consider the complete question.

An automobile firm wants to purchase either machine A or machine B. The intial cost of machines A and B are $40,000 and $50,000, respectively. The power consumption per year of machines A and B are 120 MWh and 100 MWh, respectively. The power cost is $40/MWh. The estimated operating and maintenance cost over 10 years is $30,000 for machine A and $20,000 for machine B.

First of all, we will find the amount spent on power consumption for 10 years by each machine as:

$\text{Machine A}=10\cdot 120(\$40)\\\text{Machine A}=\$48,000$